Question

$\int \frac{dx}{\left(x^2+1\right)\left(x^2+4\right)}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{\left(x^2+1\right)\left(x^2+4\right)} \\ \mathrm{Putting}{x}^2=t \\ \mathrm{Then},\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{\left(t+1\right)\left(t+4\right)} \\ \mathrm{Let}\frac{1}{\left(t+1\right)\left(t+4\right)}=\frac{A}{t+1}+\frac{B}{t+4} \\ \Rightarrow 1=A\left(t+4\right)+B\left(t+1\right) \\ \mathrm{Putting}t+4=0 \\ \Rightarrow t=-4 \\ \therefore 1=A\times 0+B\left(-3\right) \\ \Rightarrow B=-\frac{1}{3} \\ \mathrm{Putting}t+1=0 \\ \Rightarrow t=-1 \\ \therefore 1=A\left(-1+4\right)+B\times 0 \\ \Rightarrow A=\frac{1}{3} \\ \therefore \frac{1}{\left(t+1\right)\left(t+4\right)}=\frac{1}{3\left(t+1\right)}-\frac{1}{3\left(t+4\right)} \\ \Rightarrow \frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{3\left(x^2+1\right)}-\frac{1}{3\left(x^2+2^2\right)} \\ \Rightarrow \int \frac{dx}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{3}\int \frac{dx}{x^2+1^2}-\frac{1}{3}\int \frac{dx}{x^2+2^2} \\ =\frac{1}{3}{\tan }^{-1}x-\frac{1}{3}\times \frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C \\ =\frac{1}{3}{\tan }^{-1}x-\frac{1}{6}{\tan }^{-1}\left(\frac{x}{2}\right)+C \end{gathered}$$