<s>[INST] 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?
 [/INST] Trigonometric substitution There is another simpler method to solve the problem 
Let $2x=3\sec\theta \implies dx=\frac{3}{2}\sec\theta \tan\theta \ d\theta$
$$\int \frac{\sqrt{4x^2-9}}{x^2}\ dx=\int \frac{\sqrt{9\sec^2\theta-9}}{\frac{9}{4}\sec^2\theta}\ \frac{3}{2}\sec\theta \tan\theta \ d\theta$$
taking positive value, 
$$=\frac{2}{3}\int \frac{3\tan\theta}{\sec\theta}\tan\theta \ d\theta$$
$$=2\int \frac{\tan^2 \theta}{\sec\theta}\ d\theta$$
$$=2\int \frac{\sec^2 \theta-1}{\sec\theta}\ d\theta$$
$$=2\int (\sec\theta-\cos\theta)\ d\theta$$
$$=2\left(\int \sec\theta\ d\theta-\int \cos\theta\ d\theta\right)$$
$$=2\left(\ln\left|\sec\theta+\tan\theta\right| -\sin\theta\right)+c$$
substituting $\sec\theta=\frac{2x}{3}$, one should get $$=2\ln|2x+\sqrt{4x^2-9}|-\frac{\sqrt{4x^2-9}}{x}+C$$
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