Answer
$$
\int \frac{\sqrt{9 x^{2}+4}}{x^{2}} d x=-\frac{\sqrt{9 x^{2}+4}}{x}+3 \ln \left(3 x+\sqrt{9 x^{2}+4}\right)+C
$$
Work Step by Step
$$
\int \frac{\sqrt{9 x^{2}+4}}{x^{2}} d x
$$
If we make the substitution
$$
u=3 x\quad d u=3 d x
$$
and we look at the Table of Integrals , we see that the closest entry is number $24$ with $a=2:$
$$
\begin{aligned}
\int \frac{\sqrt{9 x^{2}+4}}{x^{2}} d x &=\int \frac{\sqrt{u^{2}+4}}{u^{2} / 9}\left(\frac{1}{3} d u\right) \quad\left[\begin{array}{c}
u=3 x, \\
d u=3 d x
\end{array}\right] \\
&=3 \int \frac{\sqrt{4+u^{2}}}{u^{2}} d u\\
& \stackrel{24}{=} 3\left[-\frac{\sqrt{4+u^{2}}}{u}+\ln \left(u+\sqrt{4+u^{2}}\right)\right]+C \\
&=-\frac{3 \sqrt{4+9 x^{2}}}{3 x}+3 \ln \left(3 x+\sqrt{4+9 x^{2}}\right)+C\\
&=-\frac{\sqrt{9 x^{2}+4}}{x}+3 \ln \left(3 x+\sqrt{9 x^{2}+4}\right)+C
\end{aligned}
$$
