#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 53

$\tan ^{-1} x-\frac{1}{\sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x$

Explanation:

Let

$I=\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x$

Let $x^{2}= y$

\begin{aligned} &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)}=\frac{1}{(y+1)(y+2)}=\frac{A}{y+1}+\frac{B}{y+2} \\ &\frac{1}{(y+1)(y+2)}=\frac{(y+2) A+(y+1) B}{(y+1)(y+2)} \\ &1=(y+2) A+(y+1) B \end{aligned}                                               (1)

Put $y= -1$

Equation (1)

\begin{aligned} &1=A+0 \\ &A=1 \end{aligned}

Put $y= -2$

Equation (1)

\begin{aligned} &1=0+(-1) B \\ &B=-1 \\ &\frac{1}{(y+1)(y+2)}=\frac{1}{y+2}-\frac{1}{y+2} \\ &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)}=\frac{1}{x^{2}+1}-\frac{1}{x^{2}+2} \end{aligned}

\begin{aligned} &I=\int \frac{1}{x^{2}+1} d x-\int \frac{1}{x^{2}+2} d x \\ &I=\tan ^{-1} x-\frac{1}{\sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}+C \end{aligned}