Question

$\int \sqrt{x^2+3x}dx$ is equal to
(where $C$ is integration constant)

Answer 1

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

We know that,
$\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log |x+\sqrt{x^2-a^2}|+C\therefore I=\frac{(x+\frac{3}{2})}{2}\sqrt{x^2+3x}-\frac{\frac{9}{4}}{2}\log |(x+\frac{3}{2})+\sqrt{x^2+3x}|+C=\frac{(2x+3)}{4}\sqrt{x^2+3x}-\frac{9}{8}\log |(x+\frac{3}{2})+\sqrt{x^2+3x}|+C$