Integrals — Class 12 Maths Solution

: Let$I = \int {\left( {4x + 2} \right)\sqrt {{x^2} + x + 1} dx}$

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

$I = 2\int {\left( {2x + 1} \right)\sqrt {{x^2} + x + 1} dx = 2\int {\sqrt t } dt}$

$= \cfrac{{2{t^{3/2}}}}{{3/2}} + C = \cfrac{4}{3}{t^{3/2}} + C = \cfrac{4}{3}{\left( {{x^2} + x + 1} \right)^{3/2}} + C$