Evaluate the following integrals:

$x=t^{2}$

$d(x)=2 t \cdot d t$

$d x=2 t \cdot d t$

Substituting $t$ and dt we get

$\Rightarrow \int \frac{2 t \cdot d t}{t^{2}+t}$

$\Rightarrow 2 \int \frac{t \cdot d t}{t^{2}+t}$

$\Rightarrow 2 \int \frac{1}{1+t} d t$

$\Rightarrow 2(\ln |1+t|)$

But $t=\sqrt{x}$

$\Rightarrow 2(\ln |1+\sqrt{x}|)+c$