$\int\sqrt{1+\frac{x^2}{9}}dx$ is equal to

(Where C is an arbitrary constant)

$\frac{1}{3}\left[\frac{9}{2}\log|x + \sqrt{x^2+9}|+\frac{x}{2}\sqrt{x^2+9}\right]+C$

The correct answer is Option (4) → $\frac{1}{3}\left[\frac{9}{2}\log|x + \sqrt{x^2+9}|+\frac{x}{2}\sqrt{x^2+9}\right]+C$

$\displaystyle \int \sqrt{1+\frac{x^{2}}{9}}\;dx \;=\; \frac{1}{3}\int \sqrt{x^{2}+9}\;dx$

$\displaystyle \int \sqrt{x^{2}+a^{2}}\;dx=\frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\ln\!\left|x+\sqrt{x^{2}+a^{2}}\right|+C$

$\Rightarrow\ \displaystyle \int \sqrt{1+\frac{x^{2}}{9}}\;dx = \frac{1}{3}\Big[\frac{x}{2}\sqrt{x^{2}+9}+\frac{9}{2}\ln|x+\sqrt{x^{2}+9}|\Big]+C$

$\displaystyle =\ \frac{x}{6}\sqrt{x^{2}+9}\;+\;\frac{3}{2}\ln\!\left|x+\sqrt{x^{2}+9}\right|\;+\;C$