Question

Integrate the function: $\sqrt{x^2+3x}$

Answer 1

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

=$\int \sqrt{x^2+3x+\frac{9}{4}-\frac{9}{4}}dx$

=$\int \sqrt{\bigg(x+\frac{3}{2}\bigg)^2-\bigg(\frac{3}{2}\bigg)^2}dx$

It is known that,$\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\mid x+\sqrt{x^2-a^2\mid}+C$

∴$I= \frac{\bigg(x+\frac{3}{2}\bigg)}{2}\sqrt{x^2+3x}-\frac{\frac{9}{4}}{2}\log \mid \bigg(x+\frac{3}{2}\bigg)+\sqrt{x^2+3x}\mid+C$

=$\frac{(2x+3)}{4}\sqrt{x^2+3x}-\frac{9}{8}\log \mid \bigg(x+\frac{3}{2}\bigg)+\sqrt{x^2+3x}\mid+C$