RD Sharma Class 12 Exercise 18.20 Indefinite Integrals Solutions Maths - Download PDF Free Online

Indefinite Integrals Exercise 18.20 Question 2.

Given: $x+\log \left|\frac{x-2}{x+3}\right|+c$
Hint: Using Partial Fraction
Explanation:
$I=\int \frac{x^{2}+x-1}{x^{2}+x-6} d x$
$\frac{x^{2}+x-1}{x^{2}+x-6}=\frac{x^{2}+x-6+5}{x^{2}+x-6}=1+\frac{5}{x^{2}+x-6}$
$=1+\frac{5}{(x-2)(x+3)}$
$\left[\begin{array}{l} x^{2}+x-6=x^{2}+3 x-2 x-6 \\ =x(x+3)-2(x+3) \\ =(x-2)(x+3) \end{array}\right]$
$\because \int \frac{x^{2}+x-1}{x^{2}+x-6} d x=\int 1 d x+5 \int \frac{1}{(x-2)(x+3)} d x$
$\because I=x+5 I_{1}$ .......................(1)
Where $I_{1}=\int \frac{1}{(x-2)(x+3)} d x$
$\frac{1}{(x-2)(x+3)}=\frac{A}{(x-2)}+\frac{B}{(x+3)}$
Multiplying by $(x-2)(x+3)$
$1=A(x+3)+B(x-2)$
Put x=3
$1=A(0)+B(-5) \Rightarrow B=\frac{-1}{5}$
Put x=2
$1=A(5)+B(0) \Rightarrow A=\frac{1}{5}$
$\therefore \frac{1}{(x-2)(x+3)}=\frac{\frac{1}{5}}{(x-2)}+\frac{-\frac{1}{5}}{(x+3)}$
$\therefore \int \frac{1}{(x-2)(x+3)} d x=\frac{1}{5} \int \frac{1}{x-2} d x+\left(\frac{-1}{5}\right) \int \frac{1}{x+3} d x$
$\begin{aligned} &=\frac{1}{5} \log |x-2|-\frac{1}{5} \log |x+3|+c \\ &=\frac{1}{5} \log \left|\frac{x-2}{x+3}\right|+c \end{aligned}$
Put in (1)
$\begin{aligned} &I=x+5\left(\frac{1}{5} \log \left|\frac{x-2}{x+3}\right|\right)+c \\ &I=x+\log \left|\frac{x-2}{x+3}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.20 Question 3.

Answer: $\frac{x}{2}+\log |x|-\frac{3}{4} \log |2 x-1|+c$
Given: $\int \frac{\left(1-x^{2}\right)}{x(1-2 x)} d x$
Hint: Using Partial fraction
Explanation:
Let$I=\int \frac{\left(1-x^{2}\right)}{x(1-2 x)} d x=\int \frac{1-x^{2}}{x-2 x^{2}} d x$
$I=\int \frac{x^{2}-1}{2 x^{2}-x} d x$
$\frac{x^{2}-1}{2 x^{2}-x}=\frac{1}{2}+\frac{\frac{1}{2} x-1}{2 x^{2}-x}$
$\therefore \int \frac{x^{2}-1}{2 x^{2}-x} d x=\int \frac{1}{2} d x+\int \frac{\frac{x}{2}-1}{2 x^{2}-x} d x$
$=\frac{1}{2} x+I_{1}$ ............(1) Where $I_{1}=\int \frac{\frac{x}{2}-1}{2 x^{2}-x} d x$
$\frac{\frac{x}{2}-1}{2 x^{2}-x}=\frac{\frac{x}{2}-1}{x(2 x-1)}=\frac{A}{x}+\frac{B}{2 x-1}$
Multiplying by $x(2 x-1)$
$\frac{x}{2}-1=A(2 x-1)+B(x)$
Putting x = 0
$-1=A(-1)+B(0) \Rightarrow A=1$
Putting x = 2
$\begin{aligned} &1-1=1(4-1)+B(2) \\ &0=3+2 B \\ &B=\frac{-3}{2} \end{aligned}$
$\begin{aligned} &\therefore \frac{\frac{x}{2}-1}{2 x^{2}-x}=\frac{1}{x}-\frac{3}{2(2 x-1)} \\ &\therefore I_{1}=\int \frac{1}{x} d x-\frac{3}{2} \int \frac{2}{2(2 x-1)} d x \\ &=\int \frac{1}{x} d x-\frac{3}{4} \int \frac{2}{(2 x-1)} d x \\ &=\log |x|-\frac{3}{4} \log |2 x-1| \end{aligned}$
Put in (1)
$\frac{1}{2} x+\log |x|-\frac{3}{4} \log |2 x-1|+c$

Indefinite Integrals Exercise 18.20 Question 4.

Answer: $10 \log |x-3|-5 \log |x-2|+c$
Given: $\int \frac{x^{2}+1}{x^{2}-5 x+6} d x$
Hint: Using partial fraction and $\int \frac{1}{x} d x$
Explanation:
$\begin{aligned} &\text { Let }\\ &I=\int \frac{x^{2}+1}{x^{2}-5 x+6} d x \end{aligned}$
$\frac{x^{2}+1}{x^{2}-5 x+6}=\frac{x^{2}+1}{(x-3)(x-2)} \quad\left[\begin{array}{l} x^{2}-5 x+6=x^{2}-3 x-2 x+6 \\ =x(x-3)-2(x-3) \\ =(x-3)(x-2) \end{array}\right]$
$\frac{x^{2}+1}{x^{2}-5 x+6}=\frac{A}{(x-3)}+\frac{B}{(x-2)}$
Multiplying by $\left ( x-3 \right )$ and $\left ( x-2\right )$
$x^{2}+1=A(x-2)+B(x-3)$
Putting x = 2
$4+1=A(0)+B(-1) \Rightarrow B=-5$
Putting x = 3
$9+1=A(3-2)+B(0) \Rightarrow A=10$
$\therefore \int \frac{x^{2}+1}{x^{2}-5 x+6} d x=10 \int \frac{1}{x-3} d x-5 \int \frac{1}{x-2} d x$
$=10 \log |x-3|-5 \log |x-2|+c$

Indefinite Integrals Exercise 18.20 Question 5.

Answer: $-\frac{25}{3} \log |x+5|+\frac{4}{3} \log |x+2|+c$
Given: $\int \frac{x^{2}}{x^{2}+7 x+10} d x$
Hint: Using Partial Fraction and $\int \frac{1}{x} d x$
Explanation:
Let $I=\int \frac{x^{2}}{x^{2}+7 x+10} d x$
$\frac{x^{2}}{x^{2}+7 x+10}=\frac{x^{2}}{x^{2}+5 x+2 x+10}=\frac{x^{2}}{x(x+5)+2(x+5)}=\frac{x^{2}}{(x+5)(x+2)}$
$\frac{x^{2}}{(x+5)(x+2)}=\frac{A}{x+5}+\frac{B}{x+2}$
Multiply by $(x+5)(x+2)$
$x^{2}=A(x+2)+B(x+5)$
Put x = -2
$4=A(0)+B(3) \Rightarrow B=\frac{4}{3}$
Put x = -5
$25=A(-3)+B(0) \Rightarrow A=-\frac{25}{3}$
$\frac{x^{2}}{(x+5)(x+2)}=\frac{\frac{-25}{3}}{x+5}+\frac{\frac{4}{3}}{x+2}$
$\therefore \int \frac{x^{2}}{(x+5)(x+2)} d x=\frac{-25}{3} \int \frac{1}{x+5} d x+\frac{4}{3} \int \frac{1}{x+2} d x$
$=\frac{-25}{3} \log |x+5|+\frac{4}{3} \log |x+2|+c$

Indefinite Integrals Exercise 18.20 Question 6.

Answer: $x+\log \left|x^{2}-x+1\right|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c$
Given: $\int \frac{x^{2}+x+1}{x^{2}-x+1} d x$
Hint: Using Partial Fraction
Explanation:
Let
$I=\int \frac{x^{2}+x+1}{x^{2}-x+1} d x$
$\frac{x^{2}+x+1}{x^{2}-x+1}=\frac{x^{2}-x+x+x+1}{x^{2}-x+1}=\frac{x^{2}-x+1}{x^{2}-x+1}+\frac{2 x}{x^{2}-x+1}$
$=1+\frac{2 x-1}{x^{2}-x+1}+\frac{1}{x^{2}-x+1}$
$\int \frac{x^{2}+x+1}{x^{2}-x+1} d x=\int 1 d x+\int \frac{2 x-1}{x^{2}-x+1} d x+\int \frac{1}{x^{2}-x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x$
$=\int 1 d x+\int \frac{2 x-1}{x^{2}-x+1} d x+\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x$
$=x+\log \left|x^{2}-x+1\right|+\frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c$
$=x+\log \left|x^{2}-x+1\right|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c$

Indefinite Integrals Exercise 18.20 Question 7.

Answer: $x-2 \log \left|x^{2}+x+2\right|+3 \tan ^{-1}(x+1)+c$
Given: $\int \frac{(x-1)^{2}}{x^{2}+2 x+2} d x$
Hint: $\text { Using } \int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x$
Explanation: Let
$I=\int \frac{(x-1)^{2}}{x^{2}+2 x+2} d x=\int \frac{x^{2}-2 x+1}{x^{2}+2 x+2} d x$
$=1-\frac{(4 x+1)}{x^{2}+2 x+2}$
$=1-2\left(\frac{2 x+\frac{1}{2}}{x^{2}+2 x+2}\right)$
$=1-2\left(\frac{2 x+2-2+\frac{1}{2}}{x^{2}+2 x+2}\right)$
$=1-2\left(\frac{2 x+2-2+\frac{1}{2}}{x^{2}+2 x+2}\right)$
$=1-2\left(\frac{2 x+2}{x^{2}+2 x+2}\right)-2\left(\frac{-\frac{3}{2}}{x^{2}+2 x+(1)^{2}-(1)^{2}+2}\right)$
$=1-2\left(\frac{2 x+2}{x^{2}+2 x+2}\right)+3\left(\frac{1}{(x+1)^{2}+1}\right)$
$\therefore \int \frac{(x-1)^{2}}{x^{2}+2 x+2} d x=\int 1 d x-2 \int \frac{2 x+2}{x^{2}+2 x+2} d x+3 \int \frac{1}{(x+1)^{2}+1} d x$
$=x-2 \log \left|x^{2}+2 x+2\right|+3 \tan ^{-1}(x+1)+c$

Indefinite Integrals Exercise 18.20 Question 8.

Answer: $\frac{x^{2}}{2}+2 x+\frac{3}{2} \log \left|x^{2}-x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c$
Given: $\int \frac{x^{3}+x^{2}+2 x+1}{x^{2}-x+1} d x$
Hint: Useing $\int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x$
Explanation:
$I=\int \frac{x^{3}+x^{2}+2 x+1}{x^{2}-x+1} d x$
$\frac{x^{3}+x^{2}+2 x+1}{x^{2}-x+1}=(x+2)+\left(\frac{3 x-1}{x^{2}-x+1}\right)$
$\therefore \int \frac{x^{3}+x^{2}+2 x+1}{x^{2}-2 x+1} d x=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-\frac{2}{3}}{x^{2}-x+1} d x$
$=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-1+1-\frac{2}{3}}{x^{2}-x+1} d x$
$=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-1}{x^{2}-x+1} d x+\frac{3}{2} \times \frac{1}{3} \int \frac{1}{x^{2}-x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x$
$=\int(x+2) d x+\frac{3}{2} \int \frac{2 x-1}{x^{2}-x+1} d x+\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x$
$=\frac{x^{2}}{2}+2 x+\frac{3}{2} \log \left|x^{2}-x+1\right|+\frac{1}{2} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c$
$=\frac{x^{2}}{2}+2 x+\frac{3}{2} \log \left|x^{2}-x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+c$

Indefinite Integrals Exercise 18.20 Question 9.

Answer: $\frac{x^{5}}{5}-\frac{4 x^{3}}{3}+20 x-40 \tan ^{-1} \frac{x}{2}+c$
Hint: Using $\int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x$
Explanation: Let
$I=\int \frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4} d x$
$\frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4}=\frac{\left(x^{2}+4-4\right)\left(x^{4}+4\right)}{x^{2}+4}=\frac{\left[\left(x^{2}+4\right)-4\right]\left(x^{4}+4\right)}{x^{2}+4}$
$=\frac{\left(x^{2}+4\right)\left(x^{4}+4\right)}{x^{2}+4}-\frac{4\left(x^{4}+4\right)}{x^{2}+4}$
$\therefore \int \frac{x^{2}\left(x^{4}+4\right)}{x^{2}+4} d x=\int\left(x^{4}+4\right) d x-4 \int \frac{x^{4}}{x^{2}+4} d x-16 \int \frac{1}{x^{2}+4} d x$
$=\int\left(x^{4}+4\right) d x-4 \int\left(x^{2}-\frac{4 x^{2}}{x^{2}+4}\right) d x-16 \int \frac{1}{x^{2}+4} d x$
$\left[\because \frac{x^{4}}{x^{2}+4}=x^{2}-\frac{4 x^{2}}{x^{2}+4}\right]$
$=\int\left(x^{4}+4\right) d x-4 \int x^{2} d x+16 \int \frac{x^{2}+4-4}{x^{2}+4} d x-16 \int \frac{1}{x^{2}+4} d x$
$=\int\left(x^{4}+4\right) d x-4 \int x^{2} d x+16 \int 1 d x-64 \int \frac{d x}{x^{2}+4}-16 \int \frac{d x}{x^{2}+4}$
$=\frac{x^{5}}{5}+4 x-\frac{4 x^{3}}{3}+16 x-\frac{64}{2} \tan ^{-1} \frac{x}{2}-\frac{16}{2} \tan ^{-1} \frac{x}{2}+c$
$=\frac{x^{5}}{5}+20 x-\frac{4 x^{3}}{3}-40 \tan ^{-1} \frac{x}{2}+c$
$=\frac{x^{5}}{5}-\frac{4 x^{3}}{3}+20 x-40 \tan ^{-1} \frac{x}{2}+c$

Indefinite Integrals Exercise 18.20 Question 10.

Answer: $x-3 \log \left|x^{2}+6 x+12\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c$
Given: $\int \frac{x^{2}}{x^{2}+6 x+12} d x$
Hint: Using$\int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x$
Explanation: Let
$I=\int \frac{x^{2}}{x^{2}+6 x+12} d x$
$\begin{aligned} &=\int\left(1-\left(\frac{6 x+12}{x^{2}+6 x+12}\right)\right) d x \\ &=\int 1 d x-6 \int \frac{x+2}{x^{2}+6 x+12} d x \\ &=\int 1 d x-\frac{6}{2} \int \frac{2 x+4+2-2}{x^{2}+6 x+12} d x \end{aligned}$
$\begin{aligned} &=\int 1 d x-3 \int \frac{2 x+6}{x^{2}+6 x+12} d x+6 \int \frac{1}{x^{2}+6 x+(3)^{2}-(3)^{2}+12} d x \\ &=\int 1 d x-3 \int \frac{2 x+6}{x^{2}+6 x+12}+6 \int \frac{1}{(x+3)^{2}+(\sqrt{3})^{2}} d x \end{aligned}$
$\begin{aligned} &=x-3 \log \left|x^{2}+6 x+12\right|+6 \times \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c \\ &=x-3 \log \left|x^{2}+6 x+12\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c \end{aligned}$

Indefinite Integrals Exercise 18.20 Question 11.

Answer:$x+\log \left|\frac{x}{x+1}\right|+c$
Given: $\int \frac{x^{3}+1}{x^{3}-x} d x$
Hint: using Partial Fraction and $\int \frac{1}{x} d x$
Explanation: Let
$I=\int \frac{x^{3}+1}{x^{3}-x} d x$
$\frac{x^{3}+1}{x^{3}-x}=\left(1+\frac{x+1}{x^{3}-x}\right)=1+\frac{x+1}{x\left(x^{2}-1\right)}$
$=1+\frac{x+1}{x(x-1)(x+1)}=1+\frac{1}{x(x-1)}$
$\therefore \int \frac{x^{3}+1}{x^{3}-x} d x=\int 1 d x+\int \frac{1}{x(x+1)} d x$
$=x+I_{1}$ ...................(1) Where $I_{1}=\int \frac{1}{x\left ( x+1 \right )}dx$
$\frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{(x+1)}$

Multiplying by $x\left ( x+1 \right )$

$1=A\left ( x+1 \right )+B\left ( x+1 \right )$
Putting x = -1
$1=A\left ( 0+1 \right )+B\left ( 0 \right )\Rightarrow A=1$
$\begin{aligned} &\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{(x+1)} \\ &\therefore \int \frac{1}{x(x+1)} d x=\int \frac{1}{x} d x-\int \frac{1}{x+1} d x \end{aligned}$
$=\log |x|-\log |x+1|$
Put in (1)
$\begin{aligned} I &=x+\log |x|-\log |x+1|+c \\ &=x+\log \left|\frac{x}{x+1}\right|+c \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 11

Answer: -$-\frac{1}{2} \tan ^{-2} x+3 \log |\tan x|+\frac{3}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C$
Hint: - Use substitution method to solve this integral.
Given: -$\int \frac{1}{\sin ^{3} x \cdot \cos ^{5} x} d x$
The exponent of denominator is 3+5=8. To solve this type of integral, we have to divide both numerator and denominator by $\cos ^{8 }x$ then
$\begin{aligned} &I=\int \frac{\frac{1}{\cos ^{8} x}}{\frac{\sin ^{3} x \cdot \cos ^{5} x}{\cos ^{8} x}} d x=\int \frac{\sec ^{8} x}{\frac{\sin ^{3} x}{\cos ^{3} x}} d x \\ &=\int \frac{\sec ^{8} x}{\tan ^{3} x} d x=\int \frac{\sec ^{6} x \cdot \sec ^{2} x}{\tan ^{3} x} d x=\int \frac{\left(\sec ^{2} x\right)^{3} \sec ^{2} x}{\tan ^{3} x} \end{aligned}$
$\begin{aligned} &=\int \frac{\left(1+\tan ^{2} x\right)^{3} \sec ^{2} x}{\tan ^{2} x} d x \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\because 1+\tan ^{2} x=\sec ^{2} x\right] \\ &\Rightarrow I=\int \frac{\left(1+\tan ^{6} x+3 \tan ^{2} x+3 \tan ^{4} x\right) \sec ^{2} x}{\tan ^{3} x} d x \end{aligned}$
Substitute $\tan x=t \Rightarrow \sec ^{3} x d x=d t$ then
$\begin{aligned} I &=\int \frac{\left(1+t^{6}+3 t^{2}+3 t^{4}\right)}{t^{3}} d t=\int\left(\frac{1}{t^{3}}+\frac{t^{6}}{t^{3}}+\frac{3 t^{2}}{t^{3}}+\frac{3 t^{4}}{t^{3}}\right) d t \\ &=\int\left(t^{-3}+t^{6-3}+\frac{3}{t}+3 t\right) d t=\int\left(t^{-3}+t^{3}+\frac{3}{t}+3 t\right) d t \\ &=\int t^{-3} d t+\int t^{3} d t+3 \int_{t}^{1} \frac{1}{t} d t+3 \int t d t \end{aligned}$
$=\frac{t^{-3+1}}{-3+1}+\frac{t^{3+1}}{3+1}+3 \log |t|+\frac{3 t^{1+1}}{1+1}+C \quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \frac{1}{t} d t=\log |t|+C\right]$
$\begin{aligned} &=\frac{t^{-2}}{-2}+\frac{t^{4}}{4}+3 \log |t|+\frac{3 t^{2}}{2}+C \\ &=-\frac{1}{2} t^{-2}+3 \log |t|+\frac{3 t^{2}}{2}+\frac{1}{4} t^{4}+C \\ &=-\frac{1}{2 \tan ^{2} x}+3 \log |\tan x|+\frac{3}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C \quad\quad\quad\quad[\because t=\tan x] \end{aligned}$

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