Evaluate the following integral:

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

Let, $x^{2}=y$

Then, $\frac{1}{(\mathrm{y}+1)(\mathrm{y}+2)}=\frac{\mathrm{A}}{\mathrm{y}+1}+\frac{\mathrm{B}}{\mathrm{y}+2}$

$\Rightarrow 1=A(y+2)+B(y+1)$

$\Rightarrow 1=(A+B) y+2 A+B$

On equating similar terms, we get,

$A+B=0$, and $2 A+B=1$

We get, $A=1, B=-1$

$\therefore \mathrm{I}=\int \frac{\mathrm{dx}}{\mathrm{x}^{2}+1}-\int \frac{\mathrm{dx}}{\mathrm{x}^{2}+2}$

$\Rightarrow \mathrm{I}=\tan ^{-1} \mathrm{x}-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\mathrm{x}}{\sqrt{2}}\right)+\mathrm{c}$

So, $\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x=\tan ^{-1} x-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+c$