Question

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

Answer 1

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