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?
 A: Hint: $1-\sin^2t=\cos^2t\iff9-\underbrace{9\sin^2t}_{4\,x^2}=9\cos^2t=(3\cos t)^2$
A: $$(9-4x^2)^{1/2} = \sum_{n=0}^{\infty}\binom{1/2}{n}9^{1/2-n}(-4x^2)^{n}$$
$$\dfrac{(9-4x^2)^{1/2}}{x} = \sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}\left(\dfrac{9}{x^2}\right)^{1/2-n}$$
$$\int \left(\dfrac{9}{x^2}\right)^{1/2-n}(-4)^{n} \;\mathrm{d}x =(-4)^{n}9^{1/2-n}\int {x}^{1-2n}\; \mathrm{d}x = (-4)^{n}9^{1/2-n}\left(\dfrac{x^{2-2n}}{2-2n}\right) + C$$
$$\begin{align}\int \dfrac{(9-4x^2)^{1/2}}{x} \; \mathrm{d}x &= \sum_{n=0}^{\infty}\int\binom{1/2}{n}(-4)^{n}\left(\dfrac{9}{x^2}\right)^{1/2-n} \mathrm{d}x \\ &= \sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}9^{1/2-n}\left(\dfrac{x^{2-2n}}{2-2n}\right)\\ &= \dfrac{1}{3}\sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}3^{2-2n}\left(\dfrac{x^{2-2n}}{2-2n}\right)\\ &= \dfrac{1}{6}\sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}\left(\dfrac{(3x)^{2-2n}}{1-n}\right) + c\end{align}$$
A: 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. 
A: Rewrite our integral as
$$\int 4x\frac{\sqrt{9-4x^2}}{4x^2}\,dx.$$
Let $9-4x^2=4u^2$. Then $x\,dx=-u\,du$, and we arrive at
$$\int \frac{8u^2}{4u^2-9}\,du
= \int \left( 2 + \frac{3}{2u - 3} - \frac{3}{2u + 3} \right)\ du.$$
Remark: But the answer to your question about this kind of question is probably trigonometric substitution, $2x=3\sin t$. 
