Find the integral $\int \frac{(\ln(x))^2}{x^3} \, dx$ $$\int \frac{(\ln(x))^2}{x^3} \, dx $$
Starting off with Integration by Parts
$$
\begin{align}
u = \ln(x)^2 &~~~ dv = x^{-3} \\\\
du = 2\ln(x)dx &~~~ v = \frac{x^{-2}}{-2}
\end{align}
$$
$$
\begin{align}
\int \frac{(\ln(x))^2}{x^3} &= (\ln(x))^2 \left( \frac{-x^{-2}}{2} \right) -\frac{2}{2} \int \frac{\ln(x)}{x^2}
\end{align}
$$
Integration by parts again...
$$
\begin{align}
u = \ln(x)  &\hspace{10mm} dv = \frac{1}{x^2} \\\\
du = \frac{1}{x}dx &\hspace{10mm} v = -\frac{1}{x} \\\\
-\frac{2}{2} \int \frac{\ln(x)}{x^2} &= -\ln(x) \left(\frac{1}{x} \right) + \int \frac{1}{x^2}dx \\\\ 
\end{align}
$$
Integration by parts several times $\int \frac{1}{x^2}dx \rightarrow \int \frac{1}{x} \rightarrow \ln(x) + C$
Combining everything together, I get 
$$
\int \frac{(\ln(x))^2}{x^3} = (\ln(x))^2 \left( \frac{-x^{-2}}{2} \right) -\ln(x) \left(\frac{1}{x} \right) - \ln(x) + C\\\\
= -\frac{(\ln(x))^2}{2x^2} - \frac{\ln(x)}{x} - \ln(x) + C
$$
Is there a shorter way I could have done this? I think I got the right answer, but I am not really sure either. Checking my answer via differentiate doesn't seem like a feasible test strategy and even without time constraints still seems too complex for my level right now (but then maybe this is why I need the practice)

When doing integration by parts, I recommend using this little, trusty formula (the order whether $f(x)$ or $g(x)$ comes first does not matter):
$$\int f(x)g'(x)\,dx=f(x)g(x)-\int f'(x)g(x)\,dx.$$
$$
\begin{align}
\int\frac{\ln^2{x}}{x^3}\,dx
&=-\frac{1}{2}\int\ln^2{x}\left(\frac{1}{x^2}\right)'\,dx\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}}{x^2}-\int\frac{1}{x^2}(\ln^2{x})'\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}}{x^2}-2\int\frac{\ln{x}}{x^3}\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}}{x^2}-\frac{2}{-2}\int\ln{x}\left(\frac{1}{x^2}\right)'\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}}{x^2}+\frac{\ln{x}}{x^2}-\int\frac{1}{x^2}(\ln{x})'\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}+\ln{x}}{x^2}-\int\frac{1}{x^3}\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}+\ln{x}}{x^2}-\int x^{-3}\,dx\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}+\ln{x}}{x^2}-\frac{1}{-3+1}x^{-3+1}\right)\\
&=-\frac{1}{2}\left(\frac{\ln^2{x}+\ln{x}}{x^2}+\frac{1}{2x^2}\right)\\
&=-\frac{1}{2}\left(\frac{2\ln^2{x}+2\ln{x}}{2x^2}+\frac{1}{2x^2}\right)\\
&=-\frac{2\ln^2{x}+2\ln{x}+1}{4x^2}+C.
\end{align}
$$
Wolfram Alpha check
