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?
 A: $$\int \frac{\sqrt{4x^2-9}}{x^2}dx$$
$$=-\int \sqrt{4x^2-9} \,\ d(\frac{1}{x})$$
Using by parts, $$=-\left[\frac{\sqrt{4x^2-9}}{x}-\int \frac{4}{\sqrt{4x^2-9}}dx\right]$$
$$=-\left[\frac{\sqrt{4x^2-9}}{x}-2\int \frac{d(2x)}{\sqrt{(2x)^2-3^2}}\right]$$
$$=-\left[\frac{\sqrt{4x^2-9}}{x}-2 \ln \left|2x+\sqrt{4x^2-9}\right|\right]+c$$
A: 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 $$=2\ln|2x+\sqrt{4x^2-9}|-\frac{\sqrt{4x^2-9}}{x}+C$$
A: Hint. You may integrate by parts:
$$
\int\frac{\sqrt {4x^2-9}}{x^2}dx=-\frac{\sqrt {4x^2-9}}{x}+8\int\frac1{\sqrt {4x^2-9}}dx
$$ and you may conclude easily if you know that
$$
(\text{arcsinh}\: (ax))'=\frac{a}{\sqrt {a^2x^2-1}}.
$$
A: By substitution: $\DeclareMathOperator\ach{arg\,cosh}$
Set $ x=\dfrac32\cosh t,\enspace t\ge0$, whence $\mathrm d\mkern1mu x=\dfrac32\sinh t\,\mathrm d\mkern1mu t$. With some hyperbolic trigonometry, we get 
\begin{align*}
\int \frac{\sqrt{4x^2-9}}{x^2}\,\mathrm d\mkern1mu x&=2\int \frac{\sinh t}{\cosh^2t}\sinh t\,\mathrm d\mkern1mu t =2\int \tanh^2t\,\mathrm d\mkern1mu t \\&=2\int\bigl(1-(1-\tanh^2t)\bigr)\,\mathrm d\mkern1mu t=2(t-\tanh t)\\
&=2\Biggl(\ach\frac{2x}3-\frac{\sinh\bigl(\ach\frac{2x}3\bigr)}{\frac{2x}3}\Biggr)\\
&=2\Biggl(\ach\frac{2x}3-\frac{3\sqrt{\frac{4x^2}9-1}}{2x}\Biggr)=2\Biggl(\ach\frac{2x}3-\frac{\sqrt{4x^2-9}}{2x}\Biggr)\\
&=2\ln\bigl(2x+\sqrt{4x^2-9}\bigr)-\frac{\sqrt{4x^2-9}}{x}+\text{constant}.
\end{align*}
