Evaluate the indefinite integral $\int \frac{\sqrt{9-4x^{2}}}{x}dx$ $$\int \frac{\sqrt{9-4x^{2}}}{x}dx$$
How Can I attack this kind of problem?

Let $x=\cfrac{3}{2}\sin\theta$, then $dx=\cfrac{3}{2}\cos\theta\,d\theta$.
\begin{align}
\require{cancel}
\int\frac{\sqrt{9-4x^2}}{x}\, dx&=\int\frac{\sqrt{9-4\left(\frac{3}{2}\sin\theta\right)^2}}{\cancel{\frac{3}{2}}\sin\theta}\, \cancel{\frac{3}{2}}\cos\theta\,d\theta\\
&=\int\frac{3\sqrt{1-\sin^2\theta}}{\sin\theta}\, \cos\theta\,d\theta\\
&=3\int\frac{\cos\theta}{\sin\theta}\, \cos\theta\,d\theta\\
&=3\int\frac{\cos^2\theta}{\sin\theta}\,d\theta\\
&=3\int\frac{1-\sin^2\theta}{\sin\theta}\,d\theta\\
&=3\int\frac{1}{\sin\theta}\,d\theta-3\int\sin\theta\,d\theta\\
&=3\int\frac{1}{\sin\theta}\,d\theta+3\cos\theta+C
\end{align}
The last integral can be seen here, and can be done using the substitution $u = \cos \theta$ and partial fractions.
\begin{align}
\int \frac{d\theta}{\sin \theta}
&= \int \frac{\sin \theta}{\sin^2 \theta} d\theta\\
&= \int \frac{\sin \theta}{1 - \cos^2 \theta} d\theta\\
&= \int \frac{-du}{1 - u^2}\\
&= \int \frac{du}{u^2 - 1}\\
&= \frac{1}{2}\left(\ln\ \left|1 - u\right| - \ln\ \left|1 + u\right|\right) + C_2\\
&= \frac{1}{2}\left(\ln\ \left|1 - \cos \theta\right| - \ln\ \left|1 + \cos \theta\right|\right) + C_2
\end{align}
Hope this helps Dan. 
