Evaluate the following integrals:

Solution:

Assume $x^{2}+1=t$

$\Rightarrow d\left(x^{2}+1\right)=d t$

$\Rightarrow 2 x d x=d t$

$\Rightarrow \mathrm{xdx}=\frac{\mathrm{dt}}{2}$

$x^{3}$ can be write as $x^{2} \cdot x$

$\therefore$ Now the given equation becomes

$\Rightarrow \int \frac{x^{2} \cdot x d x}{\left(x^{2}+1\right)^{3}}$

$x^{2}+1=t \Rightarrow x^{2}=t-1$

$\Rightarrow \int \frac{(t-1) d t}{2 t^{3}}$

$\Rightarrow \frac{1}{2} \int \frac{t}{t^{2}} d t-\int \frac{1}{t^{2}} d t$

$\Rightarrow \frac{1}{2} \int t^{-2} d t-\int t^{-3} d t$

$\Rightarrow \frac{1}{2}\left(-1 t^{-1}+\frac{1}{2} t^{-2}\right)+c$

But $t=\left(x^{2}+1\right)$

$\Rightarrow \frac{1}{2}\left(-1\left(x^{2}+1\right)^{-1}+\frac{1}{2}\left(x^{2}+1\right)^{-2}\right)+c$

$\Rightarrow \frac{-1}{2\left(x^{2}+1\right)}+\frac{1}{4\left(1+x^{2}\right)^{2}}+c$

$\Rightarrow \frac{-4\left(1+x^{2}\right)^{2}+2\left(1+x^{2}\right)}{8\left(1+x^{2}\right)^{3}}+c$