Question

$\int \sqrt{1+\frac{x^2}{9}}dx$ is equal to
(where $C$ is integration constant)

• A. $\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}\ln |x+\sqrt{x^2+9}|+C$
• B. $\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}{\sin }^{-1}\left(\frac{x}{3}\right)+C$
• C. $\frac{x}{6}\sqrt{x^2+9}-\frac{3}{2}\ln |x+\sqrt{x^2+9}|+C$
• D. $\frac{x}{6}\sqrt{x^2+9}+\frac{1}{2}\ln |x+\sqrt{x^2+9}|+C$

Answer 1

The correct option is A $\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}\ln |x+\sqrt{x^2+9}|+C$

$I=\int \sqrt{1+\frac{x^2}{9}}dx=\frac{1}{3}\int \sqrt{9+x^2}dx=\frac{1}{3}\int \sqrt{(3{)}^2+x^2}dx$
We know that, $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln |x+\sqrt{x^2+a^2}|+C\therefore I=\frac{1}{3}[\frac{x}{2}\sqrt{x^2+9}+\frac{9}{2}\ln |x+\sqrt{x^2+9}|]+C=\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}\ln |x+\sqrt{x^2+9}|+C$