Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{2xdx}{\left(x^2+1\right)\left(x^2+3\right)} \\ \mathrm{Putting}{x}^2=t \\ \Rightarrow 2xdx=dt \\ \therefore I=\int \frac{dt}{\left(t+1\right)\left(t+3\right)} \\ \mathrm{Let}\frac{1}{\left(t+1\right)\left(t+3\right)}=\frac{A}{t+1}+\frac{B}{t+3} \\ \Rightarrow \frac{1}{\left(t+1\right)\left(t+3\right)}=\frac{A\left(t+3\right)+B\left(t+1\right)}{\left(t+1\right)\left(t+3\right)} \\ \Rightarrow 1=A\left(t+3\right)+B\left(t+1\right) \\ \mathrm{Putting}t+3=0 \\ \Rightarrow t=-3 \\ 1=A\times 0+B\left(-3+1\right) \\ \Rightarrow B=-\frac{1}{2} \\ \mathrm{Putting}t+1=0 \\ \Rightarrow t=-1 \\ 1=A\left(-1+3\right)+B\left(-1+1\right) \\ \Rightarrow 1=A\times 2+B\times 0 \\ \Rightarrow A=\frac{1}{2} \\ \mathrm{Then}, \\ I=\frac{1}{2}\int \frac{dt}{t+1}-\frac{1}{2}\int \frac{dt}{t+3} \\ =\frac{1}{2}\log \left|t+1\right|-\frac{1}{2}\log \left|t+3\right|+C \\ =\frac{1}{2}\log \left|\frac{t+1}{t+3}\right|+C \\ =\frac{1}{2}\log \left|\frac{x^2+1}{x^2+3}\right|+C \end{gathered}$$