Careers360 Logo
School
Search Colleges, Exams, Schools & more
RD Sharma Class 12 Exercise 18.30 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.30 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 10:32 AM IST

The RD Sharma class 12 chapter 18 exercise 18.30 answers are trusted by the students on a large scale. Mathematics is a subject where many students lose marks due to lack of good practice. Therefore, it is essential for every student to own the RD Sharma Class 12 Exercise 18.30 reference book to perform well in their public exams. The 18th chapter of the class 12 mathematics syllabus consists of 31 exercises, ex 18.1 to ex 18.31. This counts to hundreds of questions that is present in this single chapter. RD Sharma solutions When it comes to ex 18.30, the difficulty of the questions would be higher as the exercises reach the end of the chapter.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.30

Indefinite Integrals exercise 18.30 question 1

Answer:
13log⁡|x+1|+53log⁡|x−2|+C
Hint:
To solve the given integration, we use partial fraction method
Given:
∫2x+1(x+1)(x−2)dx
Explanation:
Let
I=∫2x+1(x+1)(x−2)dx
Expressing the function in terms of partial fraction as following
2x+1(x+1)(x−2)=A(x+1)+B(x−2) [Using N(x)(ax+b)(cx+d)=Aax+b+Bcx+d]
2x+1(x+1)(x−2)=A(x−2)+B(x+1)(x+1)(x−2)2x+1=A(x−2)+B(x+1)2x+1=(A+B)x+(−2A+B)
On comparing the coefficient of x and constant terms we get
2=A+B (1)
−2A+B=1 (2)
Now, subtract equation (2) from (1)
A+B=2−−2A+B=13A=1A=13
Putting the value of A in equation (1)
13+B=2B=2−13B=53
Now
2x+1(x+1)(x−2)=13(x+1)+53(x−2)I=∫(13(x+1)+53(x−2))dx=∫13(x+1)dx+∫53(x−2)dx
=13log⁡|x+1|dx+53log⁡|x−2|+C [Using ∫1ax+bdx=1alog⁡|ax+b|+C]

Indefinite Integrals exercise 18.30 question 3

Answer:
x+log⁡|x−2x+3|+C
Hint:
To solve the given integration, we use partial fraction method
Given:
∫x2+x−1x2+x−6dx
Explanation:
Let
I=∫x2+x−1x2+x−6dx
I=∫(x2+x−6)+5x2+x−6dx [Adding and subtracting 5 in numerator]
∫(1+5x2+x−6)dxI=∫dx+∫5x2+x−6dxI=x+∫5x2+x−6dx (1)
Let
I1=∫5(x+3)(x−2)dx[x2+x−6=(x+3)(x−2)]
Now express the function in terms of partial fraction
5(x+3)(x−2)=Ax+3+Bx−2[N(x)(ax+b)(cx+d)=Aax+b+Bcx+d]
5(x+3)(x−2)=A(x−2)+B(x+3)(x+3)(x−2)5=A(x−2)+B(x+3)5=(A+B)x+(−2A+3B)
On comparing the coefficient we get
A+B=0(2)−2A+3B=5(3)A=−B

Equation (3)

2B+3B=55B=5B=1A=−1
Now
I1=∫(−1x+3+1x−2)dxI1=−∫1x+3dx+∫1x−2dx
I1=−log⁡|x+3|+log⁡|x−2|+C[∫1ax+bdx=1alog⁡|ax+b|]I1=log⁡|x−2x+3|+C
Putting the value of I1 in equation (1)
I=x+log⁡|x−2x+3|+C

Indefinite Integrals exercise 18.30 question 2

Answer:
18log⁡|x(x−4)(x−2)2|+C
Hint:
To solve the given integration, we use partial fraction method
Given:
∫1x(x−2)(x−4)dx
Explanation:
Let I=∫1x(x−2)(x−4)dx
Now express the functions in terms of partial fraction
1x(x−2)(x−4)=Ax+Bx−2+Cx−4[∵N(x)(ax−b)(cx−d)=Aax−b+Bcx−d]
1x(x−2)(x−4)=A(x−2)(x−4)+Bx(x−4)+Cx(x−2)x(x−2)(x−4)1=A(x2−6x+8)+B(x2−4x)+C(x2−2x)1=x2(A+B+C)+x(−6A−4B−2C)+8A
On comparing coefficient, we get
A+B+C=0(1)−6A−4B−2C=0(2)8A=1(3)A=18
Now equation (2)
⇒−68−4B−2C=0⇒2B+C=−38 (4)
Equation (1)
⇒18+B+C=0⇒B+C=−18 (5)
Subtract equation (5) from equation (4)
2B+C=−38−B+C=−18B=−28
Now equation (5)
⇒−28+C=−18C=−18+28C=18
Now
1x(x−2)(x−4)=(18x)+(−2)8(x−2)+18(x−4)I=∫(18x−28(x−2)+18(x−4))dxI=18∫1xdx−28∫1x−2dx+18∫dxx−4
I=18log⁡|x|−28log⁡|x−2|+18log⁡|x−4|+C[∫1ax−bdx=1alog⁡|ax−b|]I=18log⁡|x(x−4)(x−2)2|+C

Indefinite Integrals exercise 18.30 question 4

Answer:
−x+2log⁡|x−1|+3log⁡|x+2|+C
Hint:
To solve the given integration, we use partial fraction method
Given:
∫3+4x−x2(x+2)(x−1)dx
Explanation:
Let
I=∫3+4x−x2(x+2)(x−1)dx
I=∫3+4x−(x2+x−2)+x−2x2+x−2dx [Adding and subtract (x−2)in numerator]
I=∫1+5x−(x2+x−2)x2+x−2dxI=∫(1+5xx2+x−2−1)dxI=∫1+5x(x−1)(x+2)dx−∫dxI=∫1+5x(x−1)(x+2)dx−x(1)
Let
I1=∫1+5x(x−1)(x+2)dx1+5x(x−1)(x+2)=Ax−1+Bx+2[N(x)(ax+b)(cx+d)=Aax+b+Bcx+d]

1+5x(x−1)(x+2)=A(x+2)+B(x−1)(x−1)(x+2)(1+5x)=(A+B)x+(2A−B)
Comparing the corresponding coefficient
A+B=5 (2) 2A−B=1 (3) 

Adding equation (2) and (3)

A+B=5+2A−B=13A=6A=2
Now equation (2)
2+B=5B=3
Now
I1=∫(2x−1+3x+2)dxI1=2∫1x−1dx+3∫1x+2dxI1=2log⁡|x−1|+3log⁡|x+2|+C
Now putting the value of I1 in equation (1) and we get
I=−x+2log⁡|x−1|+3log⁡|x+2|+C

Indefinite Integrals exercise 18.30 question 5

Answer:
x+log⁡|x−1x+1|+C
Hint:
To solve the given integration, first we write the function in simple form and then apply the formula of integration
Given:
∫x2+1x2−1dx
Explanation:
I=∫x2+1x2−1dxI=∫x2−1+2x2−1dxI=∫(1+2x2−1)dxI=∫dx+∫2x2−1dx
I=x+2(12log⁡|x−1x+1|)+C[∫1x2−a2dx=12alog⁡|x−ax+a|]I=x+log⁡|x−1x+1|+C


Indefinite Integrals exercise 18.30 question 6

Answer:
12log⁡|x−1|−4log⁡|x−2|+92log⁡|x−3|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2(x−1)(x−2)(x−3)dx
Explanation:
I=∫x2(x−1)(x−2)(x−3)dxx2(x−1)(x−2)(x−3)=Ax−1+Bx−2+Cx−3[px2+qx+c(x−a)(x−b)(x−c)=Ax−a+Bx−b+Cx−c]
x2(x−1)(x−2)(x−3)=A(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)(x−1)(x−2)(x−3)x2=A(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)
 At x=24=0+B(1)(−1)+04=−BB=−4
 At x=39=0+0+C(2)(1)9=2CC=92
 At x=11=A(−1)(−2)+0+02A=1A=12
Now
x2(x−1)(x−2)(x−3)=12(x−1)+(−4)x−2+92(x−3)I=∫[12(x−1)−4x−2+92(x−3)]dxI=12∫1x−1dx−4∫1x−2dx+92∫1x−3dxI=12log⁡|x−1|−4log⁡|x−2|+92|x−3|+C

Indefinite Integrals exercise 18.30 question 7

Answer:
53log⁡|x+1|+56log⁡|x−2|−52log⁡|x+2|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫5x(x+1)(x2−4)dx
Explanation:
Let
I=∫5x(x+1)(x2−4)dxI=∫5x(x+1)(x−2)(x+2)dx… (applying the  formula a2−b2)
5x(x+1)(x−2)(x+2)=Ax+1+Bx−2+Cx+25x(x+1)(x−2)(x+2)=A(x−2)(x+2)+B(x+1)(x+2)+C(x+1)(x−2)(x+1)(x−2)(x+2)5x=A(x−2)(x+2)+B(x+1)(x+2)+C(x+1)(x−2)
 At x=25(2)=0+B(3)(4)+010=12BB=1012B=56
 At x=−25(−2)=0+0+C(−1)(−4)−10=4CC=−104 At x=−1
5(−1)=A(−3)(1)+0+0−5=−3AA=53
5x(x+1)(x−2)(x+2)=53(x+1)+56(x−2)−52(x+2)
I=∫(53(x+1)+56(x−2)−52(x+2)]dxI=53∫1x+1dx+56∫1x−2dx−52∫1x+2dxI=53log⁡|x+1|+56log⁡|x−2|−52log⁡|x+2|+C

Indefinite Integrals exercise 18.30 question 8

Answer:
log⁡|x2−1x|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2+1x(x2−1)dx
Explanation:
Let
I=∫x2+1x(x2−1)dxI=∫(x2−1)+2x(x2−1)dx [Add and subtract 1]
I=∫(1x+2x(x2−1))dxI=∫1xdx+2∫1x(x−1)(x+1)dx…(applying the formula a2−b2)
I=log⁡|x|+2I1 (1)
Where
I1=∫1x(x−1)(x+1)dx1x(x−1)(x+1)=Ax+Bx−1+Cx+11x(x−1)(x+1)=A(x−1)(x+1)+Bx(x+1)+Cx(x−1)x(x−1)(x+1)1=A(x−1)(x+1)+B(x)(x+1)+C(x)(x−1)
 At x=01=A(−1)(1)+0+0A=−1
 At x=11=0+B(1)(2)+01=2BB=12
 At x=−11=0+0+C(−1)(−2)1=2CC=12
1(x−1)(x+1)x=−1x+12(x−1)+12(x+1)I1=∫[−1x+12(x−1)+12(x+1)]dxI1=−∫1xdx+12∫1x−1dx+12∫1x+1dxI1=−log⁡|x|+12log⁡|x−1|+12log⁡|x+1|+C
Equation (1)
I=log⁡|x|−2log⁡|x|+log⁡|x−1|+log⁡|x+1|+C
I=−log⁡|x|+log⁡|(x−1)(x+1)|+CI=log⁡|x2−1x|+C

Indefinite Integrals exercise 18.30 question 9

Answer:
−110log⁡|x−1|+52log⁡|x+1|−125log⁡|2x+3|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫2x−3(x−1)(x+1)(2x+3)
Explanation:
Let
I=∫2x−3(x−1)(x+1)(2x+3)…(applying the formula a2−b2)
2x−3(x2−1)(2x+3)=2x−3(x−1)(x+1)(2x+3)
2x−3(x−1)(x+1)(2x+3)=Ax−1+Bx+1+C(2x+3)2x−3(x−1)(x+1)(2x+3)=A(x+1)(2x+3)+B(x−1)(2x+3)+C(x−1)(x+1)(x−1)(x+1)(2x+3)2x−3=A(x+1)(2x+3)+B(x−1)(2x+3)+C(x−1)(x+1)
 At x=12×1−3=A(2)(5)+B(0)+C(0)−1=10 AA=−110

 At x=−12×−1−3=A(0)+B(−2)(1)+0−5=−2BB=52
 At x=−322×(−32)−3=A(0)+B(0)+C(−32+1)(−32−1)−6=C(54)C=−245
2x−3(x−1)(x+1)(2x+3)=−110(x−1)+52(x+1)−245(2x+3)


I=−110∫1x−1x+52∫1x+1dx−245∫12x+3dxI=−110log⁡|x−1|+52log⁡|x+1|−125log⁡|2x+3|+CI=−110log⁡|x−1|+52log⁡|x+1|−125log⁡|2x+3|+C



Indefinite Integrals exercise 18.30 question 10

Answer:
x+12log⁡|x−1|−8log⁡|x−2|+272log⁡|x−3|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x3(x−1)(x−2)(2x+3)dx
Explanation:
Let
I=∫x3(x−1)(x−2)(2x+3)dxI=∫x3x3−6x2+11x−6dxI=∫(x3−6x2+11x−6)−(−6x2+11x−6)x3−6x+11x−6dx [Add and subtract−6x2+11x+6]
I=∫(1+6x2−11x+6x3−6x+11x−6)dxI=∫dx+∫6x2−11x+6(x−1)(x−2)(x−3)dxI=x+I1 (1)
Where
I1=∫6x2−11x+6(x−1)(x−2)(x−3)dx6x2−11x+6(x−1)(x−2)(x−3)=A(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)(x−1)(x−2)(x−3)6x2+6−11x=A(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)
 At x=26(4)+6−11(2)=A(0)+B(1)(−1)+C(0)30−22=−B8=−BB=−8
 At x=16(1)−11(1)+6=A(−1)(−2)+B(0)+C(0)1=2AA=12
 At x=36(9)−11(3)+6=A(0)+B(0)+C(2)(1)27=2CC=272
6x2−11x+6(x−1)(x−2)(x−3)=12(x−1)−8x−2+272(x−3)I1=12∫1x−1dx−8∫1x−2dx+272∫1x−3dxI1=12log⁡|x−1|−8log⁡|x−2|+272log⁡|x−3|+C
Equation (1)
I=x+12log⁡|x−1|−8log⁡|x−2|+272log⁡|x−3|+C

Indefinite Integrals exercise 18.30 question 11

Answer:
log⁡|(sin⁡x+2)4(sin⁡x+1)2|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫sin⁡2x(1+sin⁡x)(2+sin⁡x)dx
Explanation:
Let
I=∫sin⁡2x(1+sin⁡x)(2+sin⁡x)dxI=∫2sin⁡xcos⁡x(1+sin⁡x)(2+sin⁡x)dx[sin⁡2A=2sin⁡Acos⁡A]
Let
sin⁡x=ycos⁡xdx=dyI=∫2ydy(1+y)(2+y)
2y(1+y)(2+y)=A1+y+B2+y2y(1+y)(2+y)=A(2+y)+B(1+y)(1+y)(2+y)2y=(2A+B)+y(A+B)
Comparing coefficient
2=A+B (1)
2A+B=0 (2)
Subtract equation (1) from equation (2)
A=−2
Equation (1)
2=−2+BB=4
Now
2y(y+1)(2+y)=−2y+1+4y+2I=−2∫1y+1dy+4∫1y+2dyI=−2log⁡|y+1|+4log⁡|y+2|+C
I=log⁡|(y+2)4(y+1)2|+CI=log⁡|(sin⁡x+2)4(sin⁡x+1)2|+C[∵y=sin⁡x]

Indefinite Integrals exercise 18.30 question 12

Answer:
12log⁡|x2+1x2+3|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫2x(x2+1)(x2+3)dx
Explanation:
Let
I=∫2x(x2+1)(x2+3)dx
Let
x2=y2xdx=dyI=∫dy(y+1)(y+3)
1(y+1)(y+3)=Ay+1+By+31(y+1)(y+3)=A(y+3)+B(y+1)(y+1)(y+3)
1=A(y+3)+B(y+1) (1)
 At y=−3 equation (1) becomes 1=0+(−2)BB=1−2
 At y=−1 equation (1) becomes 1=2A+0A=12
1(y+1)(y+3)=12(y+1)−12(y+3)I=12∫1y+1dy−12∫1y+3dy
I=12log⁡|y+1|−12log⁡|y+3|+CI=12log⁡|y+1y+3|+CI=12log⁡|x2+1x2+3|+C

Indefinite Integrals exercise 18.30 question 13

Answer:
12log⁡|log⁡xlog⁡x+2|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1xlog⁡x(2+log⁡x)dx
Explanation:
I=∫1xlog⁡x(2+log⁡x)dx
Let
log⁡x=y1x=dyI=∫dyy(y+2)
Now
1y(y+2)=Ay+By+21y(y+2)=A(y+2)+By(y+2)y1=A(2+y)+By1=2A+(A+B)y
Comparing the coefficient
2A=1A=12A+B=0B=−AB=−AB=−12
1y(y+2)=12y−12(y+2)I=12∫1ydy−12∫dyy+2
I=12log⁡|y|−12log⁡|y+2|+CI=12log⁡|yy+2|+CI=12log⁡|log⁡xlog⁡x+2|+C

Indefinite Integrals exercise 18.30 question 14

Answer:
35log⁡|x+2|+15log⁡|x2+1|+15tan−1⁡x+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2+x+1(x2+1)(x+2)dx
Explanation:
Let
I=∫x2+x+1(x2+1)(x+2)dxx2+x+1(x2+1)(x+2)=Ax+2+Bx+Cx2+1
x2+x+1(x2+1)(x+2)=A(x2+1)+(Bx+C)(x+2)(x+2)(x2+1)x2+x+1=A(x2+1)+Bx2+2Bx+Cx+2Cx2+x+1=x2(A+B)+(2B+C)x+(A+2C)
Comparing the coefficient
A+B=1 (1)
2B+C=1 (2)
A+2C=1 (3)
Subtract equation (3) from equation (1), we get
A+B=1−A+2C=1B−2C=0B=2C
Equation (2)
2(2C)+C=14C+C=15C=1C=15B=25
Equation (1)
A+25=1A=1−25A=35
Now
x2+x+1(x2+1)(x+2)=35(x+2)+25x+15(x2+1)
x2+x+1(x2+1)(x+2)=35(x+2)+2x+15(x2+1)
I=35∫1x+2dx+15∫2x+1x2+1dxI=35log⁡|x+2|+15∫2xx2+1dx+15∫1x2+1dx+CI=35log⁡|x+2|+15log⁡|x2+1|+15tan−1⁡x+C

Indefinite Integrals exercise 18.30 question 15

Answer:
a(a2+b)+c(a−b)(a−c)log⁡|x−a|+b2(a+1)+c(b−a)(b−c)log⁡|x−b|+c(ac+b+1)(c−a)(c−b)log⁡|x−c|+k
Hint:
To solve this integration, we use partial fraction method
Given:
∫ax2+bx+c(x−a)(x−b)(x−c)dx
Explanation:
I=∫ax2+bx+c(x−a)(x−b)(x−c)dxax2+bx+c(x−a)(x−b)(x−c)=Ax−a+Bx−b+Cx−c
ax2+bx+c(x−a)(x−b)(x−c)=A(x−b)(x−c)+B(x−a)(x−c)+C(x−a)(x−b)(x−a)(x−b)(x−c)ax2+bx+c=A(x−b)(x−c)+B(x−a)(x−c)+C(x−a)(x−b)
 For x=bab2+b2+c=0+B(b−a)(b−c)+(0)(a+1)b2+c(b−a)(b−c)=B For x=a

a3+ba+c=A(a−b)(a−c)+0+0a3+ab+c(a−b)(a−c)=A
 For x=cac2+bc+c=0+0+C(c−a)(c−b)c(ac+b+1)(c−a)(c−b)=C
ax2+bx+c(x−a)(x−b)(x−c)=a(a2+b)+c(a−b)(a−c)⋅1x−a+b2(a+1)+c(b−a)(b−c)⋅1x−b+c(ac+b+1)(c−a)(b−c)⋅1x−cI=a(a2+b)+c(a−b)(a−c)∫1x−adx+b2(a+1)+c(b−a)(b−c)∫1x−bdx+c(ac+b+1)(c−a)(c−b)∫1x−cdxI=a(a2+b)+c(a−b)(a−c)log⁡|x−a|+b2(a+1)+c(b−a)(b−c)log⁡|x−b|+c(ac+b+1)(c−a)(c−b)log⁡|x−c|+k


Indefinite Integrals exercise 18.30 question 16

Answer:
12log⁡|x−1|−14log⁡|x2+1|+12tan−1⁡x+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x(x2+1)(x+1)dx
Explanation:
Let
I=∫x(x2+1)(x+1)dxx(x2+1)(x−1)=Ax−1+Bx+Cx2+1
x(x2+1)(x−1)=A(x2+1)+(Bx+C)(x−1)(x−1)(x2+1)x=A(x2+1)+Bx2−Bx+Cx−Cx=(A+B)x2+(C−B)x+A−C
Comparing the coefficient
A+B=0 (1)
A=−B
C−B=1 (2)
A−C=0 (3)
A=C
Equation (2)
A+A=12A=1A=12B=−12C=12
x(x2+1)(x−1)=12(x−1)+−12x+12(x2+1)=12(x−1)+1−x2(x2+1)
I=12∫1x−1dx+12∫1−xx2+1dxI=12∫1x−1dx+12∫1x2+1dx−14∫2xx2+1dxI=12log⁡|x−1|+12tan−1⁡x−14log⁡|x2+1|+C

Indefinite Integrals exercise 18.30 question 17

Answer:
16log⁡|x−1|−12log⁡|x+1|+13log⁡|x+2|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1(x−1)(x+1)(x+2)dx
Explanation:
Let
I=∫1(x−1)(x+1)(x+2)dx1(x−1)(x+2)(x+1)=Ax−1+Bx+1+Cx+2
1(x−1)(x+2)(x+1)=A(x+1)(x+2)+B(x−1)(x+2)+C(x−1)(x+1)(x−1)(x+2)(x+1)1=A(x+1)(x+2)+B(x−1)(x+2)+C(x−1)(x+1)

 At x=11=A(2)(3)+B(0)+C(0)1=6AA=16
 At x=−11=A(0)+B(−2)(1)+C(0)1=−2BB=−12
 At x=−21=A(0)+B(0)+C(−3)(−1)1=3CC=13
1(x−1)(x+1)(x+2)=16(x−1)−12(x+1)+13(x+2)I=16∫1x−1dx−12∫1x+1dx+13∫1x+2dxI=16log⁡|x−1|−12log⁡|x+1|+13log⁡|x+2|+C

Indefinite Integrals exercise 18.30 question 18

Answer:
−25tan−1⁡x2+35tan⁡x3+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2(x2+4)(x2+9)dx
Explanation:
Let
I=∫x2(x2+4)(x2+9)dxx2=(A+Bx)(x2+9)+(Cx+D)(x2+4)x2=x3(B+C)+x2(A+D)+x(9B+4C)+(9A+4D)
Comparing the coefficient
B+C=0 (1)
B=−C
A+D=1 (2)
9B+4C=0 (3)
9A+4D=1 (4)
Equation (3)
9(−C)+4C=0−9C+4C=05C=0C=0B=0
Equation (2)
A=1−D
Equation (4)
9(1−D)+4D=09−9D+4D=09=5DD=95A=1−95A=−45
Now
x2(x2+4)(x2+9)=−45(x2+4)+95(x2+9)I=−45∫1x2+4dx+95∫1x2+9dx
I=−45(12)tan−1⁡x2+95(13)tan−1⁡x3+CI=−25tan−1⁡x2+35tan−1⁡x3+C



Indefinite Integrals exercise 18.30 question 19

Answer:
log⁡|(x2−1)3x|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫5x2−1x(x−1)(x+1)dx
Explanation:
Let
I=∫5x2−1x(x−1)(x+1)dx
5x2+1x(x−1)(x+1)=Ax+Bx−1+Cx+15x2+1=A(x−1)(x+1)+B(x)(x+1)+C(x)(x−1) At x=15(1)+1=A(0)+B(1)(2)+C(0)6=2BB=3
 At x=−15(1)+1=A(0)+B(0)+C(−1)(−2)6=2CC=3
 At x=05(0)+1=A(−1)(1)+0+01=−AA=−1
5x2−1x(x−1)(x+1)=−1x+3x−1+3x+1I=−∫1xdx+3∫1x−1dx+3∫1x+1dx=−log⁡|x|+3log⁡|x−1|+3log⁡|x+1|+C
Using the formulas,(x+y)(x−y)=x2−y2&log⁡x+log⁡y=log⁡(xy)
I=log⁡|(x2−1)3x|+C

Indefinite Integrals exercise 18.30 question 20

Answer:
log⁡|x2(x−2)(x+2)2|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2+6x−8x3−4xdx
Explanation:
Let
I=∫x2+6x−8x3−4xdxI=∫x2+6x−8x(x2−4)dx
I=∫x2+6x−8x(x−2)(x+2)dx….[x2−y2=(x+y)(x−y)]x2+6x−8x(x−2)(x+2)=Ax+Bx−2+Cx+2x2+6x−8=A(x−2)(x+2)+Bx(x+2)+Cx(x−2)
 At x=00+6(0)−8=A(−2)(2)+0+0−8=−4AA=2
 At x=24+6(2)−8=B(2)(4)8=8BB=1
 At x=−24+6(−2)−8=A(0)+B(0)+C(−2)(−4)−16=8CC=−2
Now
x2+6x−8x(x−2)(x+2)=2x+1x−2−2x+2I=2∫1xdx+∫1x−2dx−2∫1x+2dxI=2log⁡|x|+log⁡|x−2|−2log⁡|x+2|+CI=log⁡|x2(x−2)(x+2)2|+C

Indefinite Integrals exercise 18.30 question 21

Answer:
16log⁡|(1+x)6(−1+x)2(2x+1)5|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2+1(2x+1)(x2−1)dx
Explanation:
Let
I=∫x2+1(2x+1)(x2−1)dxI=∫x2+1(2x+1)(x−1)(x+1)dx…[x2−y2=(x+y)(x−y)]x2+1(2x+1)(x−1)(x+1)=A2x+1+Bx−1+Cx+1x2+1=A(x−1)(x+1)+B(2x+1)(x+1)+C(x−1)(2x+1)
 At x=11+1=A(0)+B(3)(2)+C(0)2=6BB=13
B=13 At x=−11+1=A(0)+B(0)+C(−1)(−2)2=2CC=1
 At x=−1214+1=A(−12+1)(−12−1)+B(0)+C(0)54=A(−34)A=−53
x2+1(2x+1)(x−1)(x+1)=−53(2x+1)+13(x−1)+1(x+1)I=−53∫12x−1dx+13∫1x−1dx+∫1x+1dx
I=13[−52log⁡|2x+1|+log⁡|x−1|+3log⁡|x+1|]+CI=16[−5log⁡|2x+1|+2log⁡|x−1|+6log⁡|x+1|]+CI=16log⁡|(x−1)2(x+1)6(2x+1)5|+C


Indefinite Integrals exercise 18.30 question 22

Answer:
log⁡|(2log⁡x+1)◻(3log⁡x+2)1/2|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1x[6(log⁡x)2+7log⁡x+2]dx
Explanation:
I=∫1x[6(log⁡x)2+7log⁡x+2]dx
Let
log⁡x=y1xdx=dyI=∫dy6y2+7y+2dxI=∫dy(3y+2)(2y+1)
1(3y+2)(2y+1)=A3y+2+B2y+11=A(2y+1)+B(3y+2)1=y(2A+3B)+(A+2B)
Comparing the coefficient
2A+3B=0 (1)
A+2B=1 (2)
Multiply equation (2) by 2 and then
Subtract equation (1) from it
2A+4B=2−2A+3B=0B=2
Equation (2)
A+2(2)=1A+4=1A=−3
Now
1(3y+2)(2y+1)=−33y+2+22y+1I=−3∫13y+2dy+2∫12y+1dyI=−3⋅13log⁡|3y+2|+2⋅12log⁡|2y+1|I=log⁡|2y+13y+2|+C
As y=log⁡x
I=log⁡|(2log⁡x+1)◻(3log⁡x+2)|+C

Indefinite Integrals exercise 18.30 question 23

Answer:
log⁡|xnxn+1|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1x(xn+1)dx
Explanation:
Let
I=∫1x(xn+1)dxI=∫1x(xn+1)⋅xn−1xn−1dxI=∫xn−1xn(xn+1)dx
Let
xn=ynxn−1dx=dyxn−1dx=dynI=∫dyny(y+1)
I=1n∫dyy(y+1)1y(y+1)=Ay+By+11=A(y+1)+B (1)
At y=0 equation (1) becomes
1=A(1)+B(0)A=1
At y=−1 equation (1) becomes

1=A(0)+B(−1)B=−11y(y+1)=1y−1y+1
Thus
I=∫1ydy−∫1y+1dyI=log⁡|y|−log⁡|y+1|+C
As y=xn
I=log⁡|xn|−log⁡|xn+1|+CI=log⁡|xnxn+1|+C∣

Indefinite Integrals exercise 18.30 question 24

Answer:
12(b2−a2)log⁡|x−bx+a|+12(a2−b2)log⁡|x−ax+b|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x(x2−a2)(x2−b2)dx
Explanation:
Let
I=∫x(x2−a2)(x2−b2)dxI=∫x(x−a)(x−b)(x+a)(x+b)dx…(x2−y2)=(x+y)(x−y)x(x−a)(x+a)(x−b)(x+b)dx=Ax−a+Bx+a+Cx−b+Dx+b
x=A(x+a)(x−b)(x+b)+B(x−a)(x−b)(x+b)+C(x−a)(x+a)(x+b)+D(x−a)(x−b)(x+a) (1)
At x=−a equation (1) becomes
−a=A(0)+B(−2a)(−a−b)(−a+b)+C(0)+(0)−a=2Ba(a+b)(b−a)B=−12(b2−a2)
At x=a equation (1) becomes
a=2A(a)(a−b)(a+b)+B(0)+C(0)+D(0)1=2A(a2−b2)A=12(a2−b2)
At x=b equation (1) becomes
b=A(0)+B(0)+C(2b)(b−a)(b+a)b=2bC(b2−a2)C=12(b2−a2)
At x=−bequation (1) becomes
−b=A(0)+B(0)+C(0)+(−2b)(−b−a)(−b+a)−b=2Db(b+a)(a−b)D=−1a2−b2
x(x−a)(x+a)(x−b)(x+b)=12(a2−b2)(x−a)+(−1)2(b2−a2)(x+a)+12(b2−a2)(x−b)−12(a2−b2)(x+b)
I=12(a2−b2)∫1x−adx+12(b2−a2)∫dxx−b−12(b2−a2)∫dxx+a−12(a2−b2)∫dxx+bI=12(a2−b2)log⁡|x−a|+12(b2−a2)log⁡|x−b|−12(b2−a2)log⁡|x+a|
I=12(a2−b2)log⁡|x−a|+12(b2−a2)log⁡|x−b|−12(b2−a2)log⁡|x+a|−12(a2−b2)log⁡|x+b|+C
I=12(a2−b2)log⁡|x−ax+b|+12(b2−a2)log⁡|x−bx+a|+C

Indefinite Integrals exercise 18.30 question 25

Answer:
−114tan−1⁡x2+835tan−1⁡x5+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2+1(x2+4)(x2+25)dx
Explanation:
Let
I=∫x2+1(x2+4)(x2+25)dxx2+1(x2+4)(x2+25)=Ax+Bx2+4+Cx+Dx2+25x2+1=(Ax+B)(x2+25)+(Cx+D)(x2+4)x2+1=x3(A+C)+x2(B+D)+x(25A+4C)+(25B+4D)
Comparing the coefficient
Coefficient of x3
A+C=0 (2)
A=−C (3)
Coefficient of x2
B+D=1 (4)
Coefficient of x
25A+4C=0
−21C+4C=0 [From the equation (3)]
−21C=0
C=0 (5)
A=0 (6)
Constant term
25B+4D=1 (7)
Multiply the equation (4) by 4 and then subtract it from equation (7)
25B+4D=1−4B+4D=421B=−3―B=−17
Equation (4)
−17+D=1D=1+17D=87(x2+1)(x2+4)(x2+25)=−17(x2+4)+87(x2+25)
Thus
I=−17∫1x2+4dx+87∫1x2+25dxI=−17⋅(12)tan−1⁡x2+87⋅(15)tan−1⁡(x5)+CI=−114tan−1⁡x2+835tan−1⁡x5+C

Indefinite Integrals exercise 18.30 question 26

Answer:
x22+log⁡|x2−1|+12log⁡|x−1x+1|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x3+x+1x2−1dx
Explanation:
Let
I=∫x3+x+1x2−1dxI=∫(x+2x+1x2−1)dxI=∫xdx+∫2xx2−1dx+∫1x2−1dxI=[x22]+log⁡|x2−1|+12log⁡|x−1x+1|+C

Indefinite Integrals exercise 18.30 question 27

Answer:
114log⁡|x+1x+3|+52(x+1)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫3x−2(x+1)2(x+3)dx
Explanation:
Let
I=∫3x−2(x+1)2(x+3)dx3x−2(x+1)2(x+3)=A(x+1)+B(x+1)2+Cx+33x−2(x+1)2(x+3)=A(x+1)(x+3)+B(x+3)+C(x+1)2(x+1)2(x+3)
3x−2=A(x2+4x+3)+B(x+3)+C(x2+1+2x)3x−2=(A+C)x2+x(4A+B+2C)+(3A+3B+C)
Comparing the coefficient of x2,xand constant term
A+C=0
A=−C (1)
4A+B+2C=3 (2)
−4C+B+2C=3 from equation (1)
B−2C=3 (3)
3A+3B+C=−2−3C+3B+C=−2 from equation (1)
3B−2C=−2 (4)
Subtract equation (3) from equation (4)
3B−2C=−2−B−2C=32B=−5―B=−52
Equation (3)
−52−2C=3−52−3=2C−112=2CC=−114A=114
3x−2(x+1)2(x+3)=114(x+1)−52(x+1)2−114(x+3)
Thus
I=114∫1x+1dx−52∫1(x+1)2dx−114∫1x+3dxI=114log⁡|x+1|−52(1−1(x+1))−114log⁡|x+3|+CI=114log⁡|x+1x+3|+52(x+1)+C

Indefinite Integrals exercise 18.30 question 28

Answer:
325log⁡|x−3x+2|−75(x−3)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫2x+1(x+2)(x−3)2dx
Explanation:
Let
I=∫2x+1(x+2)(x−3)2dx2x+1(x+2)(x−3)2=Ax+2+Bx−3+C(x−3)22x+1=A(x−3)2+B(x−3)(x+2)+C(x+2)2x+1=x2(A+B)+x(−6A−B+C)+(9A−6B+2C)
Comparing the coefficient of x2,xand constant term
A+B=0 (1)
A=−B−6A−B+C=2−6A+A+C=2 [From equation 1]
−5A+C=2 (2)
9A−6B+2C=19A+6A+2C=115A+2C=1 (3)
Multiply the equation (2) by 2 and subtract it from equation (3)
15A+2C=1−10A+2C=425A=−3A=−325B=325 [From the equation (1)]
Equation (2)
−5(−325)+C=235+C=2x=2−35C=75

2x+1(x+2)(x−3)2=−325(x+2)+325(x−3)+75(x−3)2I=−325∫1x+2dx+325∫1x−3+75∫1(x−3)2I=−325log⁡|x+2|+325log⁡|x−3|+75⋅1(−1)(x−3)+CI=325log⁡|x−3x+2|−75(x−3)+C

Indefinite Integrals exercise 18.30 question 29

Answer:
35log⁡|x−2|+25log⁡|x+3|−1x−2+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2+1(x−2)2(x+3)dx
Explanation:
Let
I=∫x2+1(x−2)2(x+3)dxx2+1(x−2)2(x+3)=Ax−2+B(x−2)2+Cx+3
x2+1=A(x−2)(x+3)+B(x+3)+C(x−2)2x2+1=A(x2+x−6)+B(x+3)+C(x2+4−4x)x2+1=(A+C)x2+(A+B−4C)x+(−6A+3B+4C)
Equating the similar terms, we get
A+C=1 (1)
A+B−4C=0 (2)
−6A+3B+4C=1 (3)
Subtract equation (1) from equation (2) and we get
B−5C=−1 (4)
Multiply equation (1) by 6 and then adding equation (3)
3B+10C=7 (5)
Multiply equation (4) by 3 and then subtract it from equation (5)
3B+10C=73B−15C=−325C=10―
C=1025C=25
Equation (4)
B−5(25)=−1B−2=−1B=1
Equation (1)
A+25=1A=1−25A=35
Now
x2+1(x−2)2(x+3)=35(x−2)+1(x−2)2+25(x+3)
Thus
I=35∫1x−2dx+∫1(x−2)2dx+25∫1x+3dxI=35log⁡|x−2|−1x−2+25log⁡|x+3|+C

Indefinite Integrals exercise 18.30 question 30

Answer:
29log⁡|x−1x+2|−13(x−1)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x(x−1)2(x+2)dx
Explanation:
Let
I=∫x(x−1)2(x+2)dxx(x−1)2(x+2)=Ax−1+B(x−1)2+Cx+2x=A(x−1)(x+2)+B(x+2)+C(x−1)2x=x2(A+C)+x(A+B−2C)+(−2A+2B+C)
Equating the similar terms
A+C=0A=−C (1)
A+B−2C=1 (2)
A+B+2A=1B+3A=1 (3)
−2A+2B+C=0−2A+2B−A=02B−3A=0 (4)
Adding equation (3) and (4)
3B=1B=13
Equation (3)
13+3A=13A=1−133A=23A=29C=−29
x(x−1)2(x+2)=29(x−1)+13(x−1)2+−29(x+2)
I=29∫dxx−1−29∫dxx+2+13∫dx(x−1)2I=29log⁡|x−1|−29log⁡|x+2|−13(x−1)+CI=29log⁡|x−1x+2|−13(x−1)+C

Indefinite Integrals Excercise 18.30 Question 31

Answer:
14log⁡|x−1|+34log⁡|x+1|+12(x+1)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2(x−1)(x+1)2dx
Explanation:
Let
I=∫x2(x−1)(x+1)2dxx2(x−1)(x+1)2=Ax+1+Bx−1+C(x+1)2
x2=A(x+1)(x−1)+B(x+1)2+C(x−1)x2=x2(A+B)+x(2B+C)+(−A+B−C)
Equating similar terms
A+B=1 (1)
2B+C=0 (2)
C=−2B−A+B−C=0 (3)
−A+B+2B=0−A+3B=0 (4)
Adding equation (1) and (4)
4B=1B=14
Equation (2)
C=−2×14C=−12
Equation (1)
A=1−14A=34
x2(x−1)(x+1)2=34(x+1)+14(x−1)−12(x+1)2I=34∫1x+1dx+14∫1x−1dx−12∫1(x+1)2dxI=34log⁡|x+1|+14log⁡|x−1|+12(x+1)+C

Indefinite Integrals Excercise 18.30 Question 32

Answer:
1x+1+log⁡|x+2|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2+x−1(x+1)2(x+2)dx
Explanation:
Let
I=∫x2+x−1(x+1)2(x+2)dxx2+x−1(x+1)2(x+2)=Ax+1+B(x+1)2+Cx+2x2+x−1=A(x+1)(x+2)+B(x+2)+C(x+1)2x2+x−1=(A+C)x2+(3A+B+2C)x+(2A+2B+C)
Equating the similar terms
A+C=13A+B+2C=12A+2B+C=−1
On solving we get
A=0,B=−1,C=1
Thus
x2+x−1(x+1)2(x+2)=0x+1+(−1)(x+1)2+1x+2I=∫−1(x+1)2x+∫1x+2dxI=1x+1+log⁡|x+2|+C

Indefinite Integrals Excercise 18.30 Question 33

Answer:
13log⁡|x|+13x−12log⁡|2x+1|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫2x2+7x−3x2(2x+1)dx
Explanation:
Let
I=∫2x2+7x−3x2(2x+1)dx2x2+7x−3x2(2x+1)=Ax+Bx2+C2x+12x2+7x−3=Ax(2x+1)+B(2x+1)+Cx22x2+7x−3=x2(2A+C)+x(A+2B)+B
Equating similar terms
2A+C=2 (1)
A+2B=7 (2)
B=−3 (3)
Equation (2)
A−6=7A=13
Equation (1)
26+C=2C=−242x2+7x−3x2(x+1)=13x−3x2−242x+1
I=13∫dxx−13∫dxx2−24∫dx2x+1I=13log⁡|x|+13x−24⋅(12)log⁡|2x+1|+CI=13log⁡|x|+13x−12log⁡|2x+1|+C

Indefinite Integrals Excercise 18.30 Question 34

Answer:
6log⁡|x|−log⁡|x+1|−9x+1+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫5x2+20x+6x3+2x2+xdx
Explanation:
Let
I=∫5x2+20x+6x3+2x2+xdxI=∫5x2+20x+6x(x+1)2dx
5x2+20x+6x(x+1)2=Ax+Bx+1+C(x+1)25x2+20x+6=A(x+1)2+B(x)(x+1)+C(x)5x2+20x+6=x2(A+B)+x(2A+B+C)+A
Equating the similar terms
5=A+B (1)
A=6 (2)
Equation (1)
5=6+BB=−12A+B+C=2012−1+C=2011+C=20C=9
5x2+20x+6x(x+1)2=6x−1x+1+9(x+1)2I=6∫dxx−∫dxx+1+9∫dx(x+1)2I=6log⁡|x|−log⁡|x+1|−9x+1+C


Indefinite Integrals Excercise 18.30 Question 35

Answer:
94log⁡|x+2|−98log⁡|x2+4|+94tan−1⁡x2+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫18(x+2)(x2+4)dx
Explanation:
Let
I=∫18(x+2)(x2+4)dx18(x+2)(x2+4)=Ax+2+Bx+Cx2+418=A(x2+4)+(Bx+C)(x+2)18=x2(A+B)+(2B+C)x+(4A+2C)
Equating the similar terms
A+B=0A=−B (1)
2B+C=0C=−2B (2)
4A+2C=18−4B−4B=18−8B=18B=−94
Equation (2)
C=−2×(−94)C=92
Equation (1)
A=94
18(x+2)(x2+4)=94(x+2)+−94x+92x2+4=94(x+2)+−9x+184(x2+4)I=94∫dxx+2−98∫2xx2+4dx+92∫1x2+4dx
94log⁡|x+2|−98log⁡|x2+4|+94tan−1⁡x2+C

Indefinite Integrals exercise 18.30 question 36

Answer:
−12log⁡|x2+1|+2tan−1⁡x+log⁡|x+2|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫5(x2+1)(x+2)dx
Explanation:
Let
I=∫5(x2+1)(x+2)dx5(x2+1)(x+2)=Ax+Bx2+1+Cx+25=(Ax+B)(x+2)+C(x2+1)5=x2(A+C)+x(2A+B)+(2B+C)
Equating the similar terms
A+C=0A=−C (1)
2A+B=0B=−2A2B+C=5−4A−A=5 [From equation (1) and equation (2)]
−5A=5A=−1
Equation (2)
B=2
Equation (1)
C=15(x2+1)(x+2)=−x+2x2+1+1x+2I=∫2−xx2+1dx+∫1x+2dxI=2∫1x2+1dx−12∫2xdxx2+1+∫1x+2dxI=2tan−1⁡x−12log⁡|x2+1|+log⁡|x+2|+C

Indefinite Integrals exercise 18.30 question 37

Answer:
−12log⁡|x+1|+14log⁡|x2+1|+12tan−1⁡x+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x(x+1)(x2+1)dx
Explanation:
Let
I=∫x(x+1)(x2+1)dxx(x+1)(x2+1)=Ax+1+Bx+Cx2+1x=A(x2+1)+(Bx+C)(x+1)x=x2(A+B)+(B+C)x+(A+C)

Equating similar terms

A+B=0B=−A (1)
A+C=0C=−A (2)
B+C=1−A−A=1−2A=1A=−12
Equation (1)
B=12
Equation (2)
C=12x(x+1)(x2+1)=−12(x+1)+12x+12x2+1=−12(x+1)+x+12(x2+1)
I=−12∫1x+1dx+12∫xx2+1dx+12∫1x2+1dxI=−12log⁡|x+1|+14log⁡|x2+1|+12tan−1⁡x+C

Indefinite Integrals exercise 18.30 question 38

Answer:
−14log⁡|x2+1|+12tan−1⁡x+12log⁡|x+1|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫11+x+x2+x3dx
Explanation:
Let
I=∫11+x+x2+x3dxI=∫1(1+x)+x2(x+1)dxI=∫dx(1+x)(1+x2)
1(1+x2)(1+x)=Ax+B1+x2+C1+x1=(Ax+B)(1+x)+C(1+x2)1=x2(A+C)+x(B+A)+(B+C)
Equating similar terms
A+C=0A=−C (1)
B+A=0A=−B (2)
B+C=1−A−A=1−2A=1A=−12
Equation (1)
C=12
Equation (2)
B=12
11+x+x2+x3=−12x+12x2+1+12x+1=−x+12(x2+1)+12(x+1)
I=−12∫xx2+1dx+12∫1x2+1dx+12∫1x+1dxI=−14log⁡|x2+1|+12tan−1⁡x+12log⁡|x+1|+C

Indefinite Integrals exercise 18.30 question 39

Answer:
12log⁡|x+1|−12(x+1)−14log⁡|x2+1|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1(x+1)2(x2+1)dx
Explanation:
Let
I=∫1(x+1)2(x2+1)dx1(x+1)2(x2+1)=A(x+1)+B(x+1)2+Cx+Dx2+1
1=A(x+1)(x2+1)+B(x2+1)+(Cx+D)(x+1)21=(A+C)x3+(A+B+2C+D)x2+(A+C+2D)x+(A+B+D)
Equating the similar terms
A+C=0C=−A (1)
A+B+2C+D=0A+B−2A+D=0 [From equation (1)]
−A+B+D=0 (2)
A+C+2D=0
2D=0 [From equation (1)]
D=0 (3)
Equation (2)
−A+B=0A=B (4)
A+B+D=1A+A+0=1 [From equation (4) and (3)]
2A=1A=12
Equation (4)
B=12
And equation (1)
C=−12
1(x+1)2(x2+1)=12(x+1)+12(x+1)2−1x2(x2+1)I=12∫1x+1dx+12∫1(x+1)2−12∫x(x2+1)dxI=12log⁡|x+1|−12(x+1)−14log⁡|x2+1|+C

Indefinite Integrals exercise 18.30 question 40

Answer:
23log⁡|x−1|−13log⁡|x2+x+1|+23tan−1⁡(2x+13)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫2xx3−1dx
Explanation:
Let
I=∫2xx3−1dxI=∫2x(x−1)(x2+x+1)dx[a3−b3=(a−b)(a2+b2+ab)]
2x(x+1)(x2+x+1)=Ax−1+Bx+Cx2+x+12x=A(x2+x+1)+(Bx+C)(x−1)2x=x2(A+B)+x(A−B+C)+(A−C)
Equating the similar terms
A+B=0A=−B (1)
A−C=0A=C (2)
A−B+C=2A+A+A=2 [From equation (1) and (2)]
3A=2A=23
Equation (1)
B=−23
Equation (2)
C=23
2x(x−1)(x2+x+1)=23(x−1)+−23x+23x2+x+1=23(x−1)+2−2x3(x2+2x+1)
I=23∫1x−1dx−13∫2x−2(x2+x+1)dxI=23∫1x−1dx−13∫2x+1x2+x+1dx−13∫−3dxx2+x+1dxI=23∫1x−1dx−13∫2x+1x2+x+1dx+∫dx(x+12)2+(32)2dx
I=23log⁡|x−1|−13log⁡|x2+x+1|+23tan−1⁡(x+1232)+C
I=23log⁡|x−1|−13log⁡|x2+x+1|+23tan−1⁡(2x+13)+C

Indefinite Integrals exercise 18.30 question 41

Answer:
13tan−1⁡x−16tan−1⁡x2+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1(x2+1)(x2+4)dx
Explanation:
Let
I=∫1(x2+1)(x2+4)dx1(x2+1)(x2+4)=Ax+Bx2+1+Cx+Dx2+41=(Ax+B)(x2+4)+(Cx+D)(x2+1)1=x3(A+C)+x2(B+D)+(4A+C)x+4B+D
Equating the similar terms
A+C=0A=−C (1)
B+D=0B=−D (2)
4A+C=0−4C+C=0 [From the equation (1)]
−3C=0C=0A=0 [From the equation (1)]
4B+D=1−4D+D=1−3D=1
D=−13 (3)
B=13 [From the equation (2)]
1(x2+1)(x2+4)=13(x2+1)+(−1)3(x2+4)I=13∫1x2+1dx−13∫1x2+4dxI=13tan−1⁡x−13⋅(12)tan−1⁡x2+CI=13tan−1⁡x−16tan−1⁡x2+C

Indefinite Integrals exercise 18.30 question 42

Answer:
23tan−1⁡(3x2)−tan−1⁡x+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2(x2+1)(3x2+4)dx
Explanation:
Let

I=∫x2(x2+1)(3x2+4)dxx2(x2+1)(3x2+4)=Ax+Bx2+1+Cx+D3x2+4x2=(Ax+B)(3x2+4)+(Cx+D)(x2+1)x2=(3A+C)x3+(3B+D)x2+(4A+C)x+(4B+D)
Equating the similar term
3A+C=03A=−C (1)
4A+C=0
4A−3A=0 [From equation (1)]
A=0 (2)
Equation (1)
C=03B+D=1 (3)
4B+D=0D=−4B (4)
Equation (3)
3B−4B=1−B=1B=−1D=4 [From equation (4)]
I=−1∫1x2+1dx+4∫13x2+4dxI=−tan−1⁡x+43∫1x2+43dx+C
I=−tan−1⁡x+43∫1x2+(23)2dx+CI=−tan−1⁡x+43(32)tan−1⁡(x23)+CI=−tan−1⁡x+23tan−1⁡(3x2)+C

Indefinite Integrals exercise 18.30 question 43

Answer:
12log⁡|x+1x−1|−4x−1+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫3x+5x3−x2−x+1dx
Explanation:
Let
I=∫3x+5x3−x2−x+1dxI=∫3x+5x2(x−1)−1(x−1)dxI=∫3x+5(x2−1)(x−1)dx
I=∫3x+5(x+1)(x−1)2dx3x+5(x+1)(x−1)2=Ax−1+B(x−1)2+Cx+13x+5=A(x−1)(x+1)+B(x+1)+C(x−1)2 (1)
Put x=1in equation (1)
3+5=A(0)+B(2)+C(0)8=2BB=4
Put x=−1 in equation (1)
−3+5=A(0)+B(0)+C(4)2=4CC=12
Put x=0in equation (1)
5=A(−1)(1)+B+C5=−A+B+C5=−A+4+121=−A+12
12=−AA=−123x+5(x+1)(x−1)2=−12(x−1)+4(x−1)2+12(x+1)
Thus
I=−12∫dxx−1+4∫dx(x−1)2+12∫dxx+1I=−12log⁡|x−1|+(−4)1x−1+12log⁡|x+1|+CI=12log⁡|x+1x−1|−4x−1+C


Indefinite Integrals exercise 18.30 question 44

Answer:
x−log⁡|x|+12log⁡|x2+1|−tan−1⁡x+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x3−1x3+xdx
Explanation:
Let
I=∫x3−1x3+xdxI=∫x3+x−x−1x3+xdxI=∫[x3+xx3+x−x+1x3+x]dx
I=∫[1−x+1x(x2+1)]dxI=∫dx−∫x+1x(x2+1)dx
x+1x(x2+1)=Ax+Bx+Cx2+1x+1=A(x2+1)+(Bx+C)xx+1=x2(A+B)+Cx+A
Equating similar terms
A+B=0A=−B (1)
C=1 (2)
A=1B=−1 [From equation (1)]
x+1x(x2+1)=1x+−x+1x2+1
Thus
I=∫dx−∫1xdx+∫xx2+1dx−∫1x2+1dxI=x−log⁡|x|+12log⁡|x2+1|−tan−1⁡x+C

Indefinite Integrals exercise 18.30 question 45

Answer:
−2log⁡|x+1|−1x+1+3log⁡|x+2|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2+x+1(x+1)2(x+2)dx
Explanation:
Let
I=∫x2+x+1(x+1)2(x+2)dxx2+x+1(x+1)2(x+2)=Ax+1+B(x+1)2+Cx+2x2+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)2…..(1)
For x=−1 equation (1) becomes
1−1+1=A(0)+B(1)+C(0)1=B
For x=−2 equation (1) becomes
4−2+1=A(0)+B(0)+C(1)C=3
For x=0 equation (1) becomes
0+0+1=A(2)+B(2)+C1=2A+2+3 [AsB=1,C=3]
1=2A+2+31=2A+52A=−4A=−2∴x2+x+1(x+1)2(x+2)=−2x+1+1(x+1)2+3x+2
Thus
I=−2∫dxx+1+∫dx(x+1)2+3∫dxx+2I=−2log⁡|x+1|−1x+1+3log⁡|x+2|+C

Indefinite Integrals exercise 18.30 question 46

Answer:
14log⁡|x4x4+1|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1x(x4+1)dx
Explanation:
Let
I=∫1x(x4+1)dxI=∫x3x4(x4+1)dx [Multiply and divide by x3]
Let
x4=y4x3dx=dyx3dx=dy4I=14∫dyy(y+1)
Now
1y(y+1)=Ay+By+11=A(y+1)+By1=(A+B)y+A
Equating similar terms
A=1A+B=0A=−BB=−1
1y(y+1)=1y+(−1)y+1I=14∫(1y)dy+14∫−1y+1dyI=14log⁡|y|+(−14)log⁡|y+1|+C
I=14log⁡|yy+1|+CI=14log⁡|x4x4+1|+C[y=x4]

Indefinite Integrals exercise 18.30 question 47

Answer:
18log⁡x−124log⁡(x3+8)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1x(x3+8)dx
Explanation:
Let
I=∫1x(x3+8)dxI=∫x2x3(x3+8)dx [Multiply and divide byx2]
Let
x3=y3x2dx=dyx2dx=dy3I=13∫dyy(y+8)1y(y+8)=Ay+By+81=A(y+8)+By1=(A+B)y+8A
Equating both side
8A=1A=18A+B=0B=−A=−181y(y+8)=18y−18(y+8)
Thus
I=13∫(18y−18(y+8))dyI=124∫1ydy−124∫1y+8dy
I=124log⁡|y|−124log⁡|y+8|+CI=124log⁡|x3|−124log⁡|x3+8|+CI=18log⁡|x|−124log⁡|x3+8|+C

Indefinite Integrals exercise 18.30 question 48

Answer:
34[log⁡|1+x2(1−x)2|+2tan−1⁡x+C]
Hint:
To solve this integration, we use partial fraction method
Given:
∫3(1−x)(1+x2)dx
Explanation:
Let
I=∫3(1−x)(1+x2)dx3(1−x)(1+x2)=A1−x+Bx+C1+x23=A(1+x2)+(Bx+C)(1−x)3=x2(A−B)+(B−C)x+A+C
Equating the similar terms
A−B=0A=B (1)
B−C=0B=C (2)
A+C=3
B+B=3 [From equation (1) and (2)]
2B=3B=32A=B=C=32∴3(1+x2)(1−x)=32(1−x)+3x+32(1+x2)
Thus
I=32∫11−xdx+32∫x1+x2dx+32∫11+x2dxI=−32log⁡|1−x|+34log⁡|1+x2|+32tan−1⁡x+CI=34[log⁡|1+x2(1−x)2|+2tan−1⁡x+C]

Indefinite Integrals exercise 18.30 question 49

Answer:
−127log⁡|−sin⁡x+1|+19(1−sin⁡x)+16(1−sin⁡x)2+127log⁡|2+sin⁡x|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫cos⁡xdx(1−sin⁡x)3(2+sin⁡x)
Explanation:
Let
I=∫cos⁡xdx(1−sin⁡x)3(2+sin⁡x)
Let
sin⁡x=ycos⁡xdx=dyI=∫dy(1−y)3(2+y)
Now
1(1−y)3(2+y)=A1−y+B(1−y)2+C(1−y)3+D2+y1=A(1−y)2(2+y)+B(1−y)(2+y)+C(2+y)+D(1−y)3
 Put y=11=A(0)+B(0)+C(3)+D(0)1=3C
C=13 (1)
 Put y=−21=A(0)+B(0)+C(0)+D(27)1=27DD=127

 Similarly A=−127,B=191(1−y)3(2+y)=−127(1−y)+19(1−y)2+13(1−y)3+127(2+y)

Thus
I=−127∫11−ydy+19∫1(1−y)2dy+13∫1(1−y)3dy+127∫12+ydy
I=−127log⁡|1−y|+19(1−y)+16(1−y)2+127log⁡|2+y|+C……….[∫1(1−a)2da=1(1−a)]
I=−127log⁡|1−sin⁡x|+19[1−sin⁡x]+16(1−sin⁡x)+127log⁡|2+sin⁡x|+C

Indefinite Integrals exercise 18.30 question 50

Answer:
−14x+78tan−1⁡(x2)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫2x2+1x2(x2+4)dx
Explanation:
Let
I=∫2x2+1x2(x2+4)dx
Now let’s separate the fraction 2x2+1x2(x2+4) through the partial fraction
Put
4A=1A=14A+B=2B=2−AB=2−AB=2−14B=74
Equating both side
2y+1y(y+4)=14y+74(y+4)=14x2+74(x2+4)
I=14∫dxx2+74∫1x2+4dxI=14⋅(1−x)+74⋅(12)tan−1⁡x2+CI=−14x+78tan−1⁡x2+C

Indefinite Integrals exercise 18.30 question 51

Answer:
log⁡|2−sin⁡x1−sin⁡x|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫cos⁡x(1−sin⁡x)(2−sin⁡x)dx
Explanation:
Let
I=∫cos⁡x(1−sin⁡x)(2−sin⁡x)dx
Let
sin⁡x=ycos⁡xdx=dyI=∫dy(1−y)(2−y)1(1−y)(2−y)=A1−y+B2−y1=A(2−y)+B(1−y)
Put y=2
1=A(0)+B(−1)B=−1
Put
y=11=A(1)+B(0)A=1
1(2−y)(1−y)=−12−y+11−yI=∫11−ydy−∫12−ydy
I=−log⁡|1−y|+log⁡|2−y|+CI=log⁡|2−y1−y|+CI=log⁡|2−sin⁡x1−sin⁡x|+C [y=sin⁡x]

Indefinite Integrals exercise 18.30 question 52

Answer:
log⁡|(x−3)7(x−2)5|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫2x+1(x−2)(x−3)dx
Explanation:
Let
I=∫2x+1(x−2)(x−3)dx2x+1(x−2)(x−3)=Ax−2+Bx−3
(2x+1)=A(x−3)+B(x−2) (1)
Put x=3
Equation (1)
6+1=BB=7
Put x=2
Equation (1)
4+1=A(−1)A=−52x+1(x−2)(x−3)=−5x−2+7x−3
I=−5∫dxx−2+7∫dxx−3I=−5log⁡|x−2|+7log⁡|x−3|+CI=log⁡|(x−3)7(x−2)5|+C

Indefinite Integrals exercise 18.30 question 53

Answer:
tan−1⁡x−12tan−1⁡x2+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1(x2+1)(x2+2)dx
Explanation:
Let
I=∫1(x2+1)(x2+2)dx
Let x2=y
1(x2+1)(x2+2)=1(y+1)(y+2)=Ay+1+By+21(y+1)(y+2)=(y+2)A+(y+1)B(y+1)(y+2)1=(y+2)A+(y+1)B (1)
Put y=−1
Equation (1)
1=A+0A=1
Put y=−2
Equation (1)
1=0+(−1)BB=−11(y+1)(y+2)=1y+2−1y+21(x2+1)(x2+2)=1x2+1−1x2+2

I=∫1x2+1dx−∫1x2+2dxI=tan−1⁡x−12tan−1⁡x2+C


Indefinite Integrals exercise 18.30 question 54

Answer:
14log⁡|x4−1x4|+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1x(x4−1)dx
Explanation:
Let
I=∫1x(x4−1)dxI=∫x3x4(x4−1)dx [Multiply and divide byx3]
Let x4=y
4x3dx=dy
I=14∫dyy(y−1)I=14∫1+y−yy(y−1)dyI=14∫y−(y−1)y(y−1)dy
I=14[∫1y−1dy+∫−1ydy]I=14[log⁡|y−1|−log⁡|y|]+CI=14log⁡|y−1y|+C
As
y=x4I=14log⁡|x4−1x4|+C

Indefinite Integrals exercise 18.30 question 55

Answer:
14log⁡|x−1x+1|−12tan−1⁡x+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1x4−1dx
Explanation:
Let
I=∫1x4−1dxI=∫1(x−1)(x+1)(x2+1)dx1(x−1)(x+1)(x2+1)=Ax−1+Bx+1+Cx+Dx2+11=A(x2+1)(x+1)+B(x−1)(x2+1)+(Cx+D)(x2−1)
Put x=−1
1=A(0)+B(−2)(2)+(Cx+D)(0)1=−4BB=−14
Put x=1
1=A(2)(2)+B(0)+(C+D)(0)1=4AA=14
Put x=0
1=A−B+D(−1)1=14+14−D1=12−DD=−12
Put x=2
1=A(5)(3)+B(1)(5)+(5C+D)(3)1=15A+5B+15C+3D1=154−54+15C−321=104−32+15C1=10−64+15C
1=44+15C1=1+15CC=01(x−1)(x+1)(x2+1)=14(x−1)−14(x+1)−12(x2+1)
Thus
I=14∫1x−1dx−14∫dxx+1−12∫1x2+1dxI=14log⁡|x−1|−14log⁡|x+1|−12tan−1⁡x+CI=14log⁡|x−1x+1|−12tan−1⁡x+C

Indefinite Integrals exercise 18.30 question 56

Answer:
log⁡|x2+1|−log⁡|x2+2|+1(x2+2)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫2x(x2+1)(x2+2)2dx
Explanation:
Let
I=∫2x(x2+1)(x2+2)2dx
Let
x2=y2xdx=dyI=∫dy(y+1)(y+2)21(y+1)(y+2)2=Ay+1+By+2+C(y+2)21=A(y+2)2+B(y+2)(y+1)+C(y+1)

Put y=−2

1=A+(0)+(0)A=1
Put y=0
1=4A+2B+C1=4+2B−11=2B+3−2=2BB=−1

1(y+1)(y+2)2=1y+1−1(y+2)−1(y+2)2I=∫dy1+y−∫dyy+2−∫dy(y+2)2I=log⁡|1+y|−log⁡|y+2|+1(y+2)+C

As y=x2
I=log⁡|x2+1|−log⁡|x2+2|+1(x2+2)+C

Indefinite Integrals exercise 18.30 question 57

Answer:
12log⁡|x−1|+14log⁡|x2+1|+12tan−1⁡x+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2(x−1)(x2+1)dx
Explanation:
Let
I=∫x2(x−1)(x2+1)dxx2(x−1)(x2+1)=Ax−1+Bx+Cx2+1x2=A(x2+1)+(Bx+C)x−Bx−Cx2=x2(A+B)+(−B+C)x+A−C

Equating both side

A−C=0A=C (1)
−B+C=0B=C (2)
A+B=1C+C=12C=1C=12A=B=C=12
x2(x−1)(x2+1)=12(x−1)+x+12(x2+1)I=12∫dxx−1+12∫xdxx2+1+12∫1x2+1dxI=12log⁡|x−1|+14log⁡|x2+1|+12tan−1⁡x+C

Indefinite Integrals exercise 18.30 question 58

Answer:
aa2−b2tan−1⁡xa+bb2−a2tan−1⁡xb+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2(x2+a2)(x2+b2)dx
Explanation:
Let
I=∫x2(x2+a2)(x2+b2)dx
x2(x2+a2)(x2+b2)=Ax+B(x2+a2)+Cx+D(x2+b2) (i)
x2=(Ax+B)(x2+b2)+(Cx+D)(x2+a2)
On comparing coefficient
x2Coefficient
1=B+D (1)
x Coefficient
0=Ab2+Ca2 (2)
Constant
0=Bb2+Da2 (3)
x3 Coefficient 0=A+C (4)
Solving (2) and (4)
Ab2+Ca2=0A+C=0C=−AAb2−Aa2=0A(b2−a2)=0A=0,C=0
B+D=1 (1)
Bb2+Da2=0 (3)
Multiply (1) by b2and subtract (3)
Bb2+Db2=b2−Bb2+Da2=0D(b2−a2)=b2―
D=b2b2−a2B=1−DB=1−b2b2−a2B=−a2b2−a2
(i) Becomes
x2(x2+a2)(x2+b2)dx=−a2b2−a2∫1x2+a2dx+b2b2−a2∫1x2+b2dx=a2a2−b2⋅1atan−1⁡xa+b2b2−a2⋅1btan−1⁡xb=aa2−b2tan−1⁡xa+bb2−a2tan−1⁡xb+C

Indefinite Integrals exercise 18.30 question 59

Answer:
118log⁡(1+sin⁡x)−12log⁡(1−sin⁡x)+49log⁡(54−sin⁡x)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1cos⁡x(5−4sin⁡x)dx
Explanation:
Let
I=∫1cos⁡x(5−4sin⁡x)dx
 Put sin⁡x=tcos⁡xdx=dtdx=1cos⁡xdt

I=∫1cos2⁡x(5−4t)=∫1(1−sin2⁡x)(5−4t)dtI=∫1(1−t2)(5−4t)dtI=14∫1(1+t)(1−t)(54−t)dt (1)

1(1+t)(1−t)(54−t)=A1+t+B1−t+C54−t
1(1+t)(1−t)(54−t)=[A(1−t)(54−t)+B(1+t)(54−t)+C(1−t2)](1+t)(1−t)(54−t)1=A(1−t)(54−t)+B(1+t)(54−t)+C(1−t2)
 Put t=11=2B(14)B=2

 Put t=−1

1=A(2)(94)
A=29
 Put t=541=C(1−2516)1=−916C
C=−169I=14∫291+tdt+14∫2(1−t)dt−14×169∫154−tdtI=118log⁡(1+t)−12log⁡(1−t)+49log⁡(54−t)+CI=118log⁡(1+sin⁡x)−12log⁡(1−sin⁡x)+49log⁡(54−sin⁡x)+C

Indefinite Integrals exercise 18.30 question 60

Answer:
110log⁡(cos⁡x−1)−12log⁡(cos+1)+25log⁡(32+cos⁡x)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1sin⁡x(3+2cos⁡x)dx
Explanation:
Let
I=∫1sin⁡x(3+2cos⁡x)dx
 Put cox=t−sin⁡xdx=dtdx=−1sin⁡xdtI=∫−1sin2⁡x(3+2t)dt=∫−1(1−cos2⁡x)(3+2t)dtI=∫1(t2−1)(3+2t)dt
I=12∫1(t−1)(t+1)(32+t)dt (1)
1(t−1)(t+1)(32+t)=At−1+Bt+1+C32+t
1(t−1)(t+1)(32+t)=A(t+1)(32+t)+B(t−1)(32+t)+C(t2−1)(t−1)(t+1)(32+t)1=A(t+1)(32+t)+B(t−1)(32+t)+C(t2−1)
 At t=−11=−2B(12)B=−1 At t=11=2A(52)
A=15 At t=−321=C(94−1)C=45
Put in (1)
I=12×15∫1t−1dt+(−1)2∫1t+1dt+12×45∫132+tdtI=110log⁡(t−1)−12log⁡(t+1)+25log⁡(32+t)+CI=log⁡(cos⁡x−1)−12log⁡(cos⁡x+1)+25log⁡(32+cos⁡x)+C

Indefinite Integrals exercise 18.30 question 61

Answer:
16log⁡(cos⁡x−1)+12log⁡(cos⁡x+1)−23log⁡(cos⁡x+12)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫1sin⁡x+sin⁡2xdx
Explanation:
Let
I=∫1sin⁡x+sin⁡2xdxI=∫1sin⁡x+2sin⁡xcos⁡xdxI=∫1sin⁡x(1+2cos⁡x)dx
Put
cos⁡x=t−sin⁡xdx=dtdx=−1sin⁡xdxI=∫−1sin2⁡x(1+2t)dtI=∫−1(1−cos2⁡x)(1+2t)dt
I=∫12(t2−1)(12+t)dt (1)
1(t−1)(t+1)(12+t)=At−1+Bt+1+C12+t (2)
1=A(t+1)(12+t)+B(t−1)(12+t)+C(t2−1)
At t=−1
1=−2B(−12)B=1
At t=1
1=2A(32)A=13
 At t=−121=C(14−1)1=C(−34)C=−43
Put in (1) using (2)
I=12⋅13∫1t−1dt+12∫1t+1dt−12⋅43∫1t+12dtI=16log⁡(t−1)+12log⁡(t+1)−23log⁡(t+12)+CI=16log⁡(cos⁡x−1)+12log⁡(cos⁡x+1)−23log⁡(cos⁡x+12)+C

Indefinite Integrals exercise 18.30 question 62

Answer:
log⁡xex1+xex+c
Hint:
To solve this integration, we use partial fraction method
Given:
∫x+1x(1+xex)dx
Explanation:
Let
I=∫x+1x(1+xex)dx
Put
1+xex=t
Differentiate w.r.t x
xex+exdx=dtex(x+1)dx=dt(x+1)dx=1exdtI=∫1xex(t)dtI=∫1(t−1)tdt (1)
Let
1t(t−1)=At+Bt−1 (2)
1=A(t−1)+B(t) At t=11=B At t=01=−AA=−1
Put in (1) using (2)
I=∫−1tdt+∫1t−1dtI=−log⁡t+log⁡(t−1)+CI=log⁡t−1t+CI=log⁡xex1+xex+C

Indefinite Integrals exercise 18.30 question 63

Answer:
x+23tan−1⁡x3−3tan−1⁡x2
Hint:
To solve this integration, we use partial fraction method
Given:
∫(x2+1)(x2+2)(x2+3)(x2+4)dx
Explanation:
Let
I=∫(x2+1)(x2+2)(x2+3)(x2+4)dxI=∫x4+3x2+2x4+7x2+12dxI=∫(1−(4x2+10)x4+7x2+12)dxI=∫1dx−∫4x2+10x4+7x2+12dxI=x−I1 (1)
Where
I1=∫4x2+10x4+7x2+12dx4x2+10x4+7x2+12=4x2+10(x2+3)(x2+4) (i)
4x2+10=(Ax+B)(x2+4)+(Cx+D)(x2+3)
Comparing the coefficient
x3 Coefficient
0=A+C (2)
x2 Coefficient
4=B+D (3)
x Coefficient
0=4A+3C (4)
Constant
10=4B+3D (5)
 Term (2) A=−C put in (4) 0=−4C+3CC=0A=0
Multiply (3) by 4 and subtract it from (5)
4B+4D=164B+3D=10D=6B+D=4B+6=4B=−2
(i) Becomes
I1=∫−2x2+3dx+6∫1x2+4dxI1=−2⋅13tan−1⁡x3+6⋅12tan−1⁡x2
I1=−23tan−1⁡x3+3tan−1⁡x2I=x−I1I=x+23tan−1⁡x3−3tan−1⁡x2+C

Indefinite Integrals exercise 18.30 question 64

Answer:
1922tan−1⁡x2−393tan−1⁡x3+674tan−1⁡x2+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫4x4+3(x2+2)(x2+3)(x2+4)dx
Explanation:
Let
I=∫4x4+3(x2+2)(x2+3)(x2+4)dx
Put x2=y
I=∫4y2+3(y+2)(y+3)(y+4)
Let
4y2+3(y+2)(y+3)(y+4)=Ay+2+By+3+Cy+44y2+3=A(y+3)(y+4)+B(y+2)(y+4)+C(y+2)(y+3)
y=−339=−By=−467=C(−2)(−1)C=672y=−219=2AA=192
I=192∫dxx2+2−39∫dxx2+3+672∫dxx2+4I=1922tan−1⁡x2−393tan−1⁡x3+674tan−1⁡x2+C

Indefinite Integrals exercise 18.30 question 65

Answer:
x22+x+12log⁡|x−1|−14log⁡|x2+1|−12tan−1⁡x+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x4(x−1)(x2+1)dx
Explanation:
Let
I=∫x4(x−1)(x2+1)dxI=∫x4x3−x2+x−1dxI=x(x3−x2+x−1)+1(x3−x2+x−1)+1(x3−x2+x−1)I=x+1+1(x−1)(x2+1)
Let,
1(x−1)(x2+1)=Ax−1+Bx+Cx2+11=A(x2+1)+(Bx+C)(x−1)
Put x=1
1=2AA=12
Put x=0
1=A−CC=A−1=−12C=−12
Put x=−1
1=2A+2B−2C1=2(A−C)+2B1=2+2B2B=−1B=−12
∫x2(x−1)(x2+1)dx=∫xdx+∫1dx+12∫1x−1dx−12∫x+1x2+1dx=x22+x+12log⁡|x−1|−14log⁡|x2+1|−12tan−1⁡x+C

Indefinite Integrals exercise 18.30 question 66

Answer:
17log⁡x−2x+2+37tan−1⁡x3+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2x4−x2−12dx
Explanation:
Let
I=∫x2x4−x2−12dx
Let x2=y
yy2−y−12=y(y−4)(y+3)y(y−4)(y+3)=Ay−4+By+3y=A(y+3)+B(y−4)
y=−3−3=−7BB=37y=44=7AA=47
∫x2x4−x2−12=47∫1x2−4dx+37∫1x2+3dx=17log⁡x−2x+2+37⋅13tan−1⁡x3+C

Indefinite Integrals exercise 18.30 question 67

Answer:
−12tan−1⁡x+14log⁡(1+x1−x)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x21−x4dx
Explanation:
 Let I=∫x21−x4dx

 Let x2=yy1−y2=y(1+y)(1−y)=A1+y+B1−yy=A(1−y)+B(1+y)
 At y=11=2BB=12
 At y=−1−1=2AA=−12I=−12∫1x2+1dx+12∫11−x2dxI=−12tan−1⁡x+12∫11−x2dx
I=−12tan−1⁡x+12∫1(1+x)(1−x)dxI=−12tan−1⁡x+12⋅12∫2+x−x(1+x)(1−x)dxI=−12tan−1⁡x+14∫(1+x)+(1−x)(1+x)(1−x)dx
I=−12tan−1⁡x+14∫11−xdx+14∫11+xdxI=−12tan−1⁡x−14log⁡(1−x)+14log⁡(1+x)+CI=−12tan−1⁡x+14log⁡(1+x1−x)+C


Indefinite Integrals exercise 18.30 question 68

Answer:
16log⁡x−1x+1+23tan−1⁡x2+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫x2x4+x2−2dx
Explanation:
 Let I=∫x2x4+x2−2dx
 Put x2=yyy2+y−2=y(y−1)(y+2)=Ay−1+By+2y=A(y+2)+B(y−1)

 At y=11=3AA=13
 At y=−2−2=−3BB=23
∫x2x4+x2−2=13∫1x2−1dx+23∫1x2+2dx=13⋅12log⁡x−1x+1+23⋅12tan−1⁡x2+CI=16log⁡x−1x+1+23tan−1⁡x2+C


Indefinite Integrals exercise 18.30 question 69

Answer:
143tan−1⁡x3+2740log⁡x−5x+5+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫(x2+1)(x2+4)(x2+3)(x2−5)dx
Explanation:
Let
I=∫(x2+1)(x2+4)(x2+3)(x2−5)dxI=∫x4+5x2+4x4−2x2−15dxI=∫(1+7x2+19x4−2x2−15)dxI=x−∫7x2+19x4−2x2−15dx
I=x−I1 (1)
Where
I1=∫7x2+19x4−2x2−15dxx2=y7y+19y2−2y−15=7y+19(y+3)(y−5)=Ay+3+By−57y+19=A(y−5)+B(y+3)
 At y=554=8BB=274
 At y=−3−2=−8AA=14
∫7x2+19x4−2x2−15=14∫1x2+3dx+274∫1x2−5dx=14⋅13tan−1⁡x3+274⋅12×5log⁡x−5x+5+C=143tan−1⁡x3+2740log⁡x−5x+5+C

Indefinite Integrals exercise 18.30 question 70

Answer:
−log⁡(1−sin⁡x)+12log⁡(1+sin2⁡x)+tan−1⁡(sin⁡x)+C
Hint:
To solve this integration, we use partial fraction method
Given:
∫2cos⁡x(1−sin⁡x)(1+sin2⁡x)dx
Explanation:
Let
I=∫2cos⁡x(1−sin⁡x)(1+sin2⁡x)dx
 Put sin⁡x=tcos⁡xdx=dtI=∫2dt(1−t)(1+t2) (1)
2(1−t)(1+t2)=A1−t+Bt+C1+t22=A(t2+1)+(Bt+C)(1−t) At t=12=2A+(B+C)(0)A=1
 At t=02=A+C2=1+CC=1 At t=−12=2A+(−B+C)(2)2=2(1)+(−B+1)(2)2=2+2(1−B)1−B=0B=1
I=∫11−tdt+∫t+1t2+1dtI=−log⁡(1−t)+12∫2tt2+1dt+∫1t2+1dtI=−log⁡(1−t)+12log⁡(1+t2)+tan−1⁡t+CI=−log⁡(1−sin⁡x)+12log⁡(1+sin2⁡x)+tan−1⁡(sin⁡x)+C


The topics in chapter 18, ex 30 includes assessing the integrals using mathematical replacements, sums regarding coordinating and its methods, joining the parts and so on. Other than the sums provided in the textbook, there are various other questions and solved answers in the RD Sharma Class 12 Solutions Chapter 18 Exercise 18.30 Indefinite Integrals reference book.

Benefits of picking the RD Sharma Mathematics solutions from the Career360 website:

  • Many professionals in the field of mathematics have contributed their work in creating the RD Sharma Class 12 Exercise 18.30 book from scratch. Therefore, there is no doubt regarding the accuracy of the answers.

  • Students can refer to these books while completing their homework and assignments. They need not wait for a teacher to clear their doubts.

  • It becomes easier for the students to recheck their homework and assignment before submission.

  • The sample question papers available with this material can be used to perform mock tests many times before the public exams.

  • The students gain clarity in the concepts that lets them face the exams with full confidence.

RD Sharma Chapter-wise Solutions

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download E-book

Frequently Asked Questions (FAQs)

1. What is the best reference guide that the class 12 students can depend on to complete their homework?

The best reference guide, Class 12 RD Sharma Chapter 18 Exercise 18.30 book helps the students anytime while doing the homework or assignments.

2. Where can the class 12 students find the RD Sharma Class 12 Solutions Chapter 18 Ex 18.30 at a low cost?

The best source website for the students to obtain the RD Sharma Class 12 Solutions Chapter 18 Ex 18.30 is the Career360 site.

3. How many solved sums for the 30th exercise for chapter 18 is available at the RD Sharma solutions book?

The RD Sharma Solutions Class 12 RD Sharma Chapter 18 Exercise 18.30 reference book consists of 70 solved sums.

4. What RD Sharma solution books help the Class 12 students achieve more marks?

The Class 12 students feel easy to attain more marks due to the many practice questions available in the RD Sharma Class 12th Exercise 18.30 book.

5. How can the students refer to the RD Sharma solution books offline?

The RD Sharma Class 12 Solutions Chapter 18 Exercise 18.30 reference book PDF can be downloaded from the Career360 website to be used offline.

Articles

Upcoming School Exams

Central Board of Secondary Education Class 12th Examination

Admit Card Date:03 February,2025 - 04 April,2025

Andhra Pradesh Secondary School Certificate Examination

Admit Card Date:03 March,2025 - 31 March,2025

Telangana State Secondary School Certificate Examination

Admit Card Date:07 March,2025 - 04 April,2025

Karnataka Secondary School Leaving Certificate Examination

Admit Card Date:10 March,2025 - 05 April,2025

National Institute of Open Schooling 10th examination

Exam Date:17 March,2025 - 01 April,2025

View All School Exams

Also Read

Study Resources, Applications and Opportunities

NEET/JEE Coaching Scholarship
NEET/JEE Coaching Scholarship
Apply

Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes

NEET Most scoring concepts
NEET Most scoring concepts
Get now

This ebook serves as a valuable study guide for NEET 2025 exam.

NEET Previous 10 Year Questions
NEET Previous 10 Year Questions
Get now

This e-book offers NEET PYQ and serves as an indispensable NEET study material.

JEE Main Important Physics formulas
JEE Main Important Physics formulas
Get now

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

JEE Main Important Chemistry formulas
Get now

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

TOEFL ® Registrations 2024
Apply

Accepted by more than 11,000 universities in over 150 countries worldwide

View All Application Forms
News and Notifications
DoE Delhi school results 2024-25 out for classes 6, 7, 8, 9, 11; steps to check score
CBSE launches parenting calendar to strengthen parent-teacher communication, address exam stress
UP government to modernise Sanskrit schools under Project Alankar
KVS Admissions 2025 lottery results for Balvatika 1, 3 out on balvatika.kvs.gov.in
NCERT module on internet addiction to promote digital literacy among students: Education ministry
Amity International School receives special citation for Atal Tinkering Labs
Odisha to provide cooked mid-day meals for Classes 9 and 10 students
Back to top