Answer
$-\displaystyle \frac{\sqrt{4+9x^{2}}}{x}+3\ln(3x+\sqrt{4+9x^{2}}+C$
Work Step by Step
The integral is of the form
24. $\displaystyle \int\frac{\sqrt{a^{2}+u^{2}}}{u^{2}}du=-\frac{\sqrt{a^{2}+u^{2}}}{u}+\ln(u+\sqrt{a^{2}+u^{2}})+C$
---
$\displaystyle \int\frac{\sqrt{9x^{2}+2^{2}}}{x^{2}}dx=\qquad\left[\begin{array}{ll}
u=3x, & du=3dx\\
x^{2}=\frac{u^{2}}{9} & dx=\frac{du}{3}
\end{array}\right]$
$=\displaystyle \int\frac{\sqrt{2^{2}+u^{2}}}{\frac{u^{2}}{9}}\frac{du}{3}=3\int\frac{\sqrt{2^{2}+u^{2}}}{u^{2}}du$
apply formula 24
$=3[-\displaystyle \frac{\sqrt{2^{2}+(3x)^{2}}}{3x}+\ln(3x+\sqrt{2^{2}+(3x)^{2}}]+C$
$=-\displaystyle \frac{\sqrt{4+9x^{2}}}{x}+3\ln(3x+\sqrt{4+9x^{2}}+C$
