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?

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*}
