Evaluate the following integrals:

Here $(4 x+2)$ can be written as $2(2 x+1)$

Now assume, $x^{2}+x+1=t$

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

$(2 x+1) d x=d t$

$\Rightarrow \int 2(2 x+1) \sqrt{x^{2}+x+1} d x$

$\Rightarrow \int 2 \sqrt{t} d t$

$\Rightarrow \int 2 t^{1 / 2} \cdot d t$

$\Rightarrow \frac{4 \mathrm{t}^{\frac{3}{2}}}{3}+\mathrm{C}$

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

$\Rightarrow \frac{4\left(x^{2}+x+1\right)^{3 / 2}}{3}+C$