Question

Integrate the function.
$\int \sqrt{x^2+3x}dx$

Answer 1

Let $I=\int \sqrt{x^2+3x}dx=\int \sqrt{x^2+3x-(\frac{3}{2}{)}^2+(\frac{3}{2}{)}^2}dx$
$=\int \sqrt{{(x+\frac{3}{2})}^2-(\frac{3}{2}{)}^2}dx\Rightarrow I=\frac{x+\frac{3}{2}}{2}\sqrt{\left(\right(x+\frac{3}{2}{)}^2-{\left(\frac{3}{2}\right)}^2)}-\frac{9}{4\times 2}log\left|(x+\frac{3}{2}+\sqrt{x^2+3x})\right|+C[\because \int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}log|x+\sqrt{x^2-a^2}|]\Rightarrow I=\frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}log|\left(\frac{2x+3}{2}\right)+\sqrt{x^2+3x}|+C$