Integrals — Class 12 Maths Solution

Let $I = \int {\sqrt {{x^2} + 3x} } dx$

$= \int {\sqrt {1 - \left( {{x^2} + 3x + \cfrac{9}{4}} \right) - \cfrac{9}{4}} dx} = \int {\sqrt {{{\left( {x + \cfrac{3}{2}} \right)}^2} - {{\left( {\cfrac{3}{2}} \right)}^2}} dx}$

$= \cfrac{{\left( {x + \cfrac{3}{2}} \right)}}{2}\sqrt {{{\left( {x + \cfrac{3}{2}} \right)}^2} - \cfrac{9}{4}} - \cfrac{9}{8}\log \left| {\left( {x + \cfrac{3}{2}} \right) + \sqrt {{{\left( {x + \cfrac{3}{2}} \right)}^2} - \cfrac{9}{4}} } \right| + C$

$= \cfrac{{2x + 3}}{4}\sqrt {{x^2} + 3x} - \cfrac{9}{8}\log \left| {x + \cfrac{3}{2} + \sqrt {{x^2} + 3x} } \right| + C$