#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 39

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

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

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

Equating the similar terms

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

\begin{aligned} &A+B+2 C+D=0 \\ &A+B-2 A+D=0 \end{aligned}                      [From equation (1)]

$-A+B+D=0$               (2)

$A+C+2D=0$

$2D=0$                               [From equation (1)]

$D=0$                                  (3)

Equation (2)

\begin{aligned} &-A+B=0 \\ &A=B \end{aligned}                                  (4)

\begin{aligned} &A+B+D=1 \\ &A+A+0=1 \end{aligned}                    [From equation (4) and (3)]

\begin{aligned} &2 A=1 \\ &A=\frac{1}{2} \end{aligned}

Equation (4)

$B=\frac{1}{2}$

And equation (1)

$C=\frac{-1}{2}$

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