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RD Sharma Class 12 Exercise 18.31 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.31 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 11:03 AM IST

RD Sharma books are considered the benchmark material for maths. The majority of CBSE schools across the country use RD Sharma books for exam references. A majority of teachers use this book for their class lectures and homework too. This is because of the comprehensive and informative nature of this book. In addition, many students prefer RD Sharma books as they contain a lot of concepts and are helpful for preparation.

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  1. RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
  2. Indefinite Integrals Excercise:18.31
  3. The benefits of RD Sharma class 12 solutions chapter 18 exercise 18.31 Solutions are:

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.31

Indefinite Integrals Exercise18.31 Question 1

Answer: The required value of the integral is,
I=13tan−1⁡(x2−13x)+c
Hint: Use the formula ∫1x2+a2dx=1atan−1⁡xa+c
Given:
∫(x2+1)(x4+x2+1)dx
Solution: The equation can be written as
I=∫1+1x2x2+1+1x2dx [ dividing x2 both denominator and denominator]
=∫1+1x2(x−1x2)2+3dx [Making the perfect square as (a+b)2 ]
Let x−1x=t
Now differentiating both side w.r.t t
(1+1x2)=dtI=∫1t2+3dt
On using standard identity we get, I=13(t3)+c
Substituting t=x−1x we get,
I=13tan−1⁡((x−1x)3)+cI=13tan−1⁡(x2−13x)+c
So, the required value of the integration is,
I=13tan−1⁡(x2−13x)+c

Indefinite Integrals Exercise18.31 Question 2

Answer: The required value of the integral is,
I=−12tan−1⁡(cot⁡θ−12cot⁡θ)−122log⁡|cot⁡θ+1−2cot⁡θcot⁡θ+1+2cot⁡θ|+c

Hint: Use the identity formula ∫1x2−a2dx=12alog⁡(x−ax+a)+c

Given:I=∫cot⁡θdθ
Solution: Suppose cot⁡θasx2
Differentiating the above equation with respect to θ.
−cosec2θdθ=2xdxdθ=−2x1+cot2⁡θdθdθ=−2x1+x4dxI=∫−2x1+x4dx….1+cot2⁡θ=cosec2θ
Re-writing the given equation as,
I=∫1+1x2+1−1x21x2+x2dx−∫1+1x2(x−1x)2+2dx−∫1−1x2(x+1x)2−2 [Making the perfect square as (a+b)2 ]
 Let, x−1x=t and x+1x=z so, I=−∫dtt2+2−∫dzz2−2

On using identity,

=−12(x−1x2)−122log⁡|x+1x−2x+1x+2|+c
The required value of the integral is,
=−12tan−1⁡(x−1x2)−122log⁡|x+1x−2x+1x+2|+c
Now, substitutex=cot⁡θinto the above equation we get,
I=−12tan−1⁡(cot⁡θ−12cot⁡θ)−122log⁡|cot⁡θ+1−2cot⁡θcot⁡θ+1+2cot⁡θ|+c
Where c is the integrating constant.

Indefinite Integrals Exercise18.31 Question 3

Answer: The required value of the integral is,
I=132(x−9x32)+c
Hint: Use the formula∫1x2+a2dx=1atan−1⁡xa+c
Given:∫(x2+9)(x4+81)dx
Solution: The equation can be written as
I=∫1+9x2x2+81x2dx
=∫1+9x2(x−9x)2+18dx
Let, x−9x=t then,
Differentiating the above equation both sides.
(1+9x2)dx=dt then the obtained equation become
∫dtt2+(32)2
Now, using the identity rule,
∫1x2+1dx=tan−1⁡(x)+c=132tan−1⁡(t32)+c
Now, re-substituting t=x−9x then,
I=132tan−1⁡(x−9x32)+c=132tan−1⁡(x2−932x)+c
Where c is the integrating constant.

Indefinite Integrals Exercise18.31 Question 4

Answer: The required value of the integral is,
I=12[13(x−1x3)−12log⁡|x+1x−1x+1x+1|]
Hint: Use the identity formula
∫1x2+1dx=1atan−1⁡xa+c and ∫dxx2−1=12alog⁡|x−ax+a|+c
Given:
∫1x4+x2+1dx
Solution: The given equation can be written as,
I=∫1x2x2+1+1x2dx [dividing x2 in both denminator and numerator]
=12∫1+1x2+1x2−1x2+1+1x2dx=12[∫1+1x2x2+1+1x2dx+∫−1+1x2(x+1x)2−1dx]=12[∫1+1x2(x−1x)2+3dx+∫−1+1x2(x+1x)2−1dx] [Making the perfect square as(a+b)2 ]
Re-writing the given equation as,
 Let, x−1x=t and x+1x=z so (1+1x2)dx=dt and (1−1x2)dx=dzI=12[∫dtt2+3−∫dzz2−1]
On using identity,
∫1x2+1dx=tan−1⁡x and ∫dzz2−1=12log⁡|z−1z+1|+cI=12[13tan−1⁡(t3)−12log⁡|z−1z+1|]
Now, re-substituting t=x−1xandz=x+1x
The required value of the integral is,
I=12[13tan−1⁡(x−1x3)−12log⁡|x+1x−1x+1x+1|]+c=123tan−1⁡(x2−1x3)−14log⁡|x2−x+1x2+x+1|+c
Where c is the integrating constant.

Indefinite Integrals Exercise18.31 Question 5

Answer: The required value of the integral is,
I=13tan−1⁡(x−1x3)−3tan−1⁡(2x2+13)+c
Hint: Use the identity formula∫1x2+1dx=1atan−1⁡xa+c
Given:I=∫x2−3x+1x4+x2+1dx
Solution: The given equation can be written as,
I=∫1−3x+1x2x2+1+1x2dx=∫1+1x2(x−1x)2+3dx−∫3xx4+x2+1dx [dividing x2 i both numerator and denominator]
[ Making the perfect square as (a+b)2 ]
Now, substitutingx−1x=tandx2=z then,
(1+1x2)dx=dtan⁡d2xdx=dzI=∫dtt2+3−32∫dzz2+z+1
We knows
∫1x2+a2dx=1atan−1⁡(xa)+C
=13tan−1⁡(t3)+C1−32∫dz(z+12)2+(32)2=13tan−1⁡(t3)+C1−32×12tan−1⁡(|z+12|32)+C2
t=x−14 and z=x2=13tan−1⁡(x−1x3)+C1−3tan−1⁡(2x2+13)+C2=13tan−1⁡(x2−13x)−3tan−1⁡(2x2+13)+C

Indefinite Integrals Exercise18.31 Question 6

Answer: The required value of the integration is,
tan−1⁡(x2−1x)+c
Hint: Use the identity formula ∫1x2+a2dx=1atan−1⁡xa+c
Given: The given integral is I=∫x2+1x4−x2+1dx
Solution:
The given equation can be written as,
I=∫x2+1x2(x4−x2+1x2)dx=∫(1+1x2)dx(x2−1+1x2)=∫(1+1x2)dxx2+1x2+1−2 [divinding x2 on both numerator and denominator ]
=∫(1+1x2)dx(x−1x)2+12 [Making the perfect square of (a+b)2 ]
Now, substitute x−1x=t then,
(1+1x2)dx=dt
Now, solving the obtained equation,
I=∫dtt2+1=tan−1⁡t+c
Re-substitute t=x−1x into the above equation.
I=tan−1⁡(x−1x)+c=tan−1⁡(x2−1x)
Thus the required value of the given integration is,
tan−1⁡(x2−1x)+c

Indefinite Integrals Exercise18.31 Question 7

Answer: The required value of the integration is,
122log⁡|x+1x−2x+1x+2|+c
Hint: Use the identity formula ∫1x2+a2dx=1atan−1⁡xa+c
Given:
I=∫x2−1x4+1dx

Solution: Given equation can be written as
I=x2−1x4+1dx
Divide numerator and denominator by x2
I=∫1−1x2x2+1x2dx=∫1−1x2x2+1x2+2−2dx=∫(1−1x2)(x+1x)2−2dx
Letx+1x=t
(1−1x2)dx=dt differentiating on both sides with respect to t,
I=∫dtt2−2=∫dt(t)2−(2)2=122log⁡|t−2t+2|+c
Re- substitutet=x+1x into the above equation.
122log⁡|x+1x−2x+1x+2|+c
Where c is an integrating constant.

Indefinite Integrals Exercise18.31 Question 8

Answer: The required value of the integral is,
13tan−1⁡(x2−13x)+c
Given:
∫x2+1x4+7x2+1dx
Hint: The given integral function can be converted into
∫dxx2+a2=1atan−1⁡xa+c
Solution: Rewrite the given equation as.
∫x2+1x4+7x2+1dx
Divide numerator and denominator by x2
I=∫1+1x2x2+1x2+7dx=∫1+1x2x2+1x2+7+2−2dx=∫1+1x2(x2+1x2−2)+9dx=∫1+1x2(x−1x)2+9dx
let(x−1x)=t⇒(1+1x2)dx=dt
Evaluating the obtained integration.
I=∫dtt2+32I=13tan−1(t3)+cI=13tan−1⁡(x−1x)+c3)+cI=13tan−1⁡(x2−13x)+c (Substituting t as (x−1x) ) 

Indefinite Integrals Exercise18.31 Question 9

Answer:
13(x2−13x)−23(2x2+13)+c
Given:
∫x2−1x4+x2+1
Hint:
The given integral function can be converted into
∫dxx2+a2=1atan−1⁡xa+c
Solution
I=∫(x−1)2x4+x2+1dx=∫x2−2x+1x4+x2+1dx=∫x2+1x4+x2+1dx+∫−2xx4+x2+1dx
 let x2=t2xdx=dtI=∫x2(1+1x2)x2(x2+1+1x2)dx+∫−dtt2+t+1 (substituting t as x2)=∫(1+1x2)(x−1x2)2+2+1dx−∫1(t+12)2−14+1dt
 let x−1x=p(1+1x2)dx=dpI=∫dpp2+3−∫dt(t+12)2+34( substituting p as x−1x)=∫1p2+(3)2dp−∫1(t+12)2+(32)2dt
Using the identity
∫dxx2+a2=1axa+c
I=13tan−1⁡p3−23tan−1⁡(t+1232)=13tan−1⁡(x−1x3)−23tan−1⁡(2x2+13) (substituting p as (x−1x) and t as x2)=13tan−1⁡(x2−13x)−23tan−1⁡(2x2+13)+c
So, the required value of the given integral is,
13(x2−13x)−23(2x2+13)+c
where c is integrating constant.

Indefinite Integrals Exercise18.31 Question 10

Answer:
125(x2−15x)−12(x2+1x)+c
Given:
∫1x4+3x2+1dx
Hint:
The given integral function can be converted into
∫dxx2+a2=1atan−1⁡xa+c
Solution
I=∫1x4+3x2+1dx=∫1x2x2+3+1x2dx(Numerator and denominator divided by x2)
=12∫(1+1x2)−(1−1x2)x2+3+1x2dx=12[∫1+1x2(x−1x)2+5dx−∫1−1x2(x+1x)2+1dx]
 Assume t=x−1x and z=x+1xdt=(1+1x2)dx and dz=(1−1x2)dx=12[∫dtt2+5−∫dzz2+1]
Using identity ∫dxx2+a2=1axa+c
I=125tan−(t5)−12tan−1⁡z
Substituting tasx−1xandzasx+1x
I=125tan−1⁡(x−1x5)−12tan−1⁡(x+1x)+c125(x2−15x)−12(x2+1x)+c
Where c is integrating constant.

Indefinite Integrals Exercise18.31 Question 11

Answer:
I=13tan−1⁡(tan2⁡x−13tan⁡x)+c
Hint:The given integral function can be converted into,
∫dxx2+a2=1atan−1⁡xa+c
Given:
I=∫1sin4⁡x+sin2⁡xcos2⁡x+cos4⁡x
Solution
I=∫1sin4⁡x+sin2⁡xcos2⁡x+cos4⁡x=∫(tan2⁡x+1)sec2⁡xtan4⁡x+tan2⁡x+1dx (dividing N and D by cos4⁡x)
=∫t2+1t4+t2+1dt where tan⁡x=t
=∫1+1t2t2+1t2+1dt (Dividing N and D by t2)
puttingt−1t=zsothat(1+1t2)dt=dzandt2+1t2=z2+2I=∫dzz2+(3)2=13tan−⁡z3+c=13tan−⁡(t2−13t)+c
The required value of the given integration is,
I=13tan−1⁡(tan2⁡x−13tan⁡x)+c
where c is an integrating constant.

Indefinite Integrals Exercise18.32 Question 12

Answer : - 13log⁡|x+2−3x+2+3|+C
Hint :- Use substitution method to solve this integral.
Given :- ∫1(x−1)x+2dx
Sol : - Let I=∫1(x−1)x+2dx
Put x+2=t2⇒dx=2tdt
I=∫1(t2−2−1)t22tdt(∵x=t2−2)=2∫1(t2−3)ttdt=2∫1t2−3dt
=2⋅123log⁡|t−3t+3|+C(∵∫1x2−a2dx=1/2alog⁡|x−ax+a|+C)=13log⁡|x+2−3x+2+3|+C(∵t=x+2) Ans.. 

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