Question

$\int \frac{1}{\left(x^2+2\right)\left(x^2+5\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{\left(x^2+2\right)\left(x^2+5\right)} \\ \mathrm{Putting}{x}^2=t \\ \therefore \frac{1}{\left(x^2+2\right)\left(x^2+5\right)}=\frac{1}{\left(t+2\right)\left(t+5\right)} \\ \mathrm{Let}\frac{1}{\left(t+2\right)\left(t+5\right)}=\frac{A}{t+2}+\frac{B}{t+5} \\ \Rightarrow \frac{1}{\left(t+2\right)\left(t+5\right)}=\frac{A\left(t+5\right)+B\left(t+2\right)}{\left(t+2\right)\left(t+5\right)} \\ \Rightarrow 1=A\left(t+5\right)+B\left(t+2\right) \\ \mathrm{Putting}t=-5 \\ \therefore 1=B\left(-5+2\right) \\ \Rightarrow B=-\frac{1}{3} \\ \mathrm{Putting}t=-2 \\ \therefore 1=A\left(-2+5\right)+B\times 0 \\ \Rightarrow A=\frac{1}{3} \\ \therefore I=\frac{1}{3}\int \frac{dx}{x^2+2}-\frac{1}{3}\int \frac{dx}{x^2+5} \\ =\frac{1}{3}\int \frac{dx}{x^2+{\left(\sqrt{2}\right)}^2}-\frac{1}{3}\int \frac{dx}{x^2+{\left(\sqrt{5}\right)}^2} \\ =\frac{1}{3\sqrt{2}}{\tan }^{-1}\left(\frac{x}{\sqrt{2}}\right)-\frac{1}{3\sqrt{5}}{\tan }^{-1}\left(\frac{x}{\sqrt{5}}\right)+C \end{gathered}$$