Evaluate the following integral:

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

Let $x^{2}+2=t \Rightarrow 2 x d x=d t$

$\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{(\mathrm{t}-1) \mathrm{t}^{2}}$

Now, let, $\frac{1}{(t-1) t^{2}}=\frac{A}{t-1}+\frac{B}{t}+\frac{C}{t^{2}}$

$\Rightarrow 1=A t^{2}+B t(t-1)+C(t-1)$

For $t=1, A=1$

For $t=0, C=-1$

For $t=-1, B=-1$

$\therefore I=\int \frac{d t}{t-1}-\int \frac{d t}{t}-\int \frac{d t}{t^{2}}$

$\Rightarrow I=\log |t-1|-\log |t|+\frac{1}{t}+c$

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