Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx}$

Put $2{x^2} + x - 3 = t$ $\Rightarrow$ $\left( {4x + 1} \right)dx = dt$

$\therefore$ $I = \int {\cfrac{{dt}}{{\sqrt t }} = \int {{t^{1/2}}} } dt = 2{t^{1/2}} + C = 2\sqrt {2{x^2} + x - 3} + C$