#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 21

$\frac{1}{6} \log \left|\frac{(1+x)^{6}(-1+x)^{2}}{(2 x+1)^{5}}\right|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

\begin{aligned} &\text { At } x=1 \\ &1+1=A(0)+B(3)(2)+C(0) \\ &2=6 B \\ &B=\frac{1}{3} \end{aligned}
\begin{aligned} &B=\frac{1}{3} \\ &\text { At } x=-1 \\ &1+1=A(0)+B(0)+C(-1)(-2) \\ &2=2 C \\ &C=1 \end{aligned}
\begin{aligned} &\text { At } x=\frac{-1}{2} \\ &\frac{1}{4}+1=A\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}-1\right)+B(0)+C(0) \\ &\frac{5}{4}=A\left(\frac{-3}{4}\right) \\ &A=\frac{-5}{3} \end{aligned}
\begin{aligned} &\frac{x^{2}+1}{(2 x+1)(x-1)(x+1)}=\frac{-5}{3(2 x+1)}+\frac{1}{3(x-1)}+\frac{1}{(x+1)} \\ &I=\frac{-5}{3} \int \frac{1}{2 x-1} d x+\frac{1}{3} \int \frac{1}{x-1} d x+\int \frac{1}{x+1} d x \end{aligned}
\begin{aligned} &I=\frac{1}{3}\left[\frac{-5}{2} \log |2 x+1|+\log |x-1|+3 \log |x+1|\right]+C \\ &I=\frac{1}{6}[-5 \log |2 x+1|+2 \log |x-1|+6 \log |x+1|]+C \\ &I=\frac{1}{6} \log \left|\frac{(x-1)^{2}(x+1)^{6}}{(2 x+1)^{5}}\right|+C \end{aligned}