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

$\frac{1}{2} \log \left|\frac{x^{2}+1}{x^{2}+3}\right|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

Let

\begin{aligned} &x^{2}=y \\ &2 x d x=d y \\ &I=\int \frac{d y}{(y+1)(y+3)} \end{aligned}

\begin{aligned} &\frac{1}{(y+1)(y+3)}=\frac{A}{y+1}+\frac{B}{y+3} \\ &\frac{1}{(y+1)(y+3)}=\frac{A(y+3)+B(y+1)}{(y+1)(y+3)} \end{aligned}

$1=A(y+3)+B(y+1)$                              (1)
\begin{aligned} &\text { At } y=-3 \text { equation }(1) \text { becomes }\\ &1=0+(-2) B\\ &B=\frac{1}{-2} \end{aligned}
\begin{aligned} &\text { At } y=-1 \text { equation (1) becomes }\\ &1=2 A+0\\ &A=\frac{1}{2} \end{aligned}
\begin{aligned} &\frac{1}{(y+1)(y+3)}=\frac{1}{2(y+1)}-\frac{1}{2(y+3)} \\ &I=\frac{1}{2} \int \frac{1}{y+1} d y-\frac{1}{2} \int \frac{1}{y+3} d y \end{aligned}
\begin{aligned} &I=\frac{1}{2} \log |y+1|-\frac{1}{2} \log |y+3|+C \\ &I=\frac{1}{2} \log \left|\frac{y+1}{y+3}\right|+C \\ &I=\frac{1}{2} \log \left|\frac{x^{2}+1}{x^{2}+3}\right|+C \end{aligned}