#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 48

$\frac{3}{4}\left[\log \left|\frac{1+x^{2}}{(1-x)^{2}}\right|+2 \tan ^{-1} x+C\right]$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{3}{(1-x)\left(1+x^{2}\right)} d x \\$

Explanation:

Let
\begin{aligned} &I=\int \frac{3}{(1-x)\left(1+x^{2}\right)} d x \\ &\frac{3}{(1-x)\left(1+x^{2}\right)}=\frac{A}{1-x}+\frac{B x+C}{1+x^{2}} \\ &3=A\left(1+x^{2}\right)+(B x+C)(1-x) \\ &3=x^{2}(A-B)+(B-C) x+A+C \end{aligned}

Equating the similar terms

\begin{aligned} &A-B=0 \\ &A=B \end{aligned}                              (1)

\begin{aligned} &B-C=0 \\ &B=C \end{aligned}                                  (2)

$A+C= 3$

$B+B=3$                           [From equation (1) and (2)]

\begin{aligned} &2 B=3 \\ &B=\frac{3}{2} \\ &A=B=C=\frac{3}{2} \\ &\therefore \frac{3}{\left(1+x^{2}\right)(1-x)}=\frac{3}{2(1-x)}+\frac{3 x+3}{2\left(1+x^{2}\right)} \end{aligned}

Thus

\begin{aligned} &I=\frac{3}{2} \int \frac{1}{1-x} d x+\frac{3}{2} \int \frac{x}{1+x^{2}} d x+\frac{3}{2} \int \frac{1}{1+x^{2}} d x \\ &I=\frac{-3}{2} \log |1-x|+\frac{3}{4} \log \left|1+x^{2}\right|+\frac{3}{2} \tan ^{-1} x+C \\ &I=\frac{3}{4}\left[\log \left|\frac{1+x^{2}}{(1-x)^{2}}\right|+2 \tan ^{-1} x+C\right] \end{aligned}