TITLE: Evaluate the indefinite integral $\int \frac{\sqrt{9-4x^{2}}}{x}dx$
QUESTION [7 upvotes]: $$\int \frac{\sqrt{9-4x^{2}}}{x}dx$$
How Can I attack this kind of problem?

REPLY [14 votes]: 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, $2x=3\sin t$.