Evaluate the following integrals:

Assume, $2 x^{2}+3 x+1=t$

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

$(4 x+3) d x=d t$

Substituting $t$ and dt in above equation we get

$\Rightarrow \int \frac{1}{\sqrt{t}} \mathrm{dt}$

$\Rightarrow \int \mathrm{t}^{-1 \backslash 2} \cdot \mathrm{dt}$

$\Rightarrow 2 \mathrm{t}^{1 \backslash 2}+\mathrm{c}$

But $t=2 x^{2}+3 x+1$

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