Use integration by parts to find the integral $\int\frac{\sqrt {4x^2-9}}{x^2}dx$ $$\int\frac{\sqrt {4x^2-9}}{x^2}dx$$ 
I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?

$$\int \frac{\sqrt{4x^2-9}}{x^2}dx$$
$$=-\int \sqrt{4x^2-9} \,\ d(\frac{1}{x})$$
Using by parts, $$=-\left[\frac{\sqrt{4x^2-9}}{x}-\int \frac{4}{\sqrt{4x^2-9}}dx\right]$$
$$=-\left[\frac{\sqrt{4x^2-9}}{x}-2\int \frac{d(2x)}{\sqrt{(2x)^2-3^2}}\right]$$
$$=-\left[\frac{\sqrt{4x^2-9}}{x}-2 \ln \left|2x+\sqrt{4x^2-9}\right|\right]+c$$
