Question:
$(4 x+2) \sqrt{x^{2}+x+1}$
Solution:
Let $x^{2}+x+1=t$
$\therefore(2 x+1) d x=d t$
$\int(4 x+2) \sqrt{x^{2}+x+1} d x$
$=\int 2 \sqrt{t} d t$
$=2 \int \sqrt{t} d t$
$=2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+\mathrm{C}$
$=\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+\mathrm{C}$