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)

$$\begin{align}
\int \frac{(\ln(x))^2}{x^3}dx & = \bigg((\ln(x))^2\times \frac{x^{-2}}{-2}\bigg)-\int\frac{2\ln(x)}{x}\times \frac{x^{-2}}{-2}dx \\
& =\bigg((\ln(x))^2\times \frac{x^{-2}}{-2}\bigg)+\int \ln(x)\times x^{-3}dx \\
& =\bigg((\ln(x))^2\times \frac{x^{-2}}{-2}\bigg)+\bigg(\ln(x)\times \frac{x^{-2}}{-2}\bigg)-\int\frac{1}{x}\times\frac{x^{-2}}{-2}dx \\
& =\bigg((\ln(x))^2\times \frac{x^{-2}}{-2}\bigg)+\bigg(\ln(x)\times \frac{x^{-2}}{-2}\bigg)+\frac{1}{2}\bigg(\frac{x^{-2}}{-2}\bigg)+C \\
& =\bigg(\frac{x^{-2}}{-2}\bigg)\bigg((\ln(x))^2+\ln(x)+\frac{1}{2}\bigg)+C
\end{align}$$
