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)

Starting off with Integration by Parts
$$
\begin{align}
u = \ln(x)^2 &~~~ dv = x^{-3} \\\\
\color{red}{du = \frac{2\ln(x)}{x}dx} &~~~ v = -\frac{x^{-2}}{2}
\end{align}
$$
(Don't forget chain rule!)
$$
\begin{align}
\int \frac{(\ln(x))^2}{x^3} &= (\ln(x))^2 \left( \frac{-x^{-2}}{2} \right) \color{red}{+\int \frac{\ln(x)}{x^3}dx}
\end{align}
$$
(Notice that integration by parts is $uv - \int vdu$)
Now we want to figure out 
$$ \int \frac{\ln(x)}{x^3}dx$$
Let
$$u = \ln(x), \ \ dv = \frac{dx}{x^3},$$
$$du = \frac{1}{x}dx, \ \ v = -\frac{x^{-2}}{2}.$$ Then we have
$$ \int \frac{\ln(x)}{x^3}dx = -\frac{x^{-2}}{2} \ln(x) + \frac{1}{2}\int\frac{1}{x^3}dx.$$
Now
$$ \int \frac{1}{x^3}dx = -\frac{x^{-2}}{2} + C,$$
so putting it all back together we have
$$ -\frac{\ln(x)^2}{2x^2} - \frac{\ln(x)}{2x^2} - \frac{1}{4x^2} + C.$$
