Question

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

• A. $\frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}\log |x+\frac{3}{2}+\sqrt{x^2+3x}|+C$
• B. $\frac{x}{2}\sqrt{x^2+3x}-\frac{9}{8}\log |x+\frac{3}{2}+\sqrt{x^2+3x}|+C$
• C. $\frac{x}{2}\sqrt{x^2+3x}-\frac{9}{8}{\sin }^{-1}\left(\frac{2x+3}{2}\right)+C$
• D. $\frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}{\sin }^{-1}\left(\frac{2x+3}{2}\right)+C$

Answer 1

The correct option is A $\frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}\log |x+\frac{3}{2}+\sqrt{x^2+3x}|+C$

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

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