How do I solve $\int_1^{2}\frac{(\ln{x})^2}{x^3} dx$? $\int_1^{2}\frac{(\ln{x})^2}{x^3} dx$
This seemed like an integration by parts problem, so I used it and got:
$\int_1^{2}\frac{(\ln{x})^2}{x^3} dx = [-\frac{1}{2}x^{-2}(\ln{x})^2 | _{1}^{2}] - \int_1^{2}\frac{2\ln{x}}{x^3}dx$
But I can't get anywhere with the $\int_1^{2}\frac{2\ln{x}}{x^3}dx$ part, making me think I didn't choose the right approach. Am I missing something simple?
 A: Hint :
Put $\ln x=t$ and hence $dx=e^t dt$
The integral then changes to $$\int_{0}^{\ln 2} t^2e^{-2t}dt$$
Now apply repeated integration by parts with $u=t^2$ and $dv=e^{-2t}dt$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\int_{1}^{2}{\ln^{2}\pars{x} \over x^{3}}\,\dd x =
\left.\partiald[2]{}{\nu}\int_{1}^{2}x^{\nu}\,\right\vert_{\ \nu\ =\ -3} =
\partiald[2]{}{\nu}\pars{2^{\nu + 1} - 1\over \nu + 1}_{\ \nu\ =\ -3}
\\[5mm] = &\
\bbx{3 - 2\ln\pars{2} - 2\ln^{2}\pars{2} \over 16} \approx 0.0408
\end{align}
A: Alternatively, you can use integration by parts two times to get rid of the natural logarithm in the numerator.
$$\begin{align*}\int\limits_1^2dx\,\frac {\log^2 x}{x^3} & =\left.-\frac {\log^2x}{2x^2}\right|_1^2+\int\limits_1^2dx\,\frac {\log x}{x^3}\\ & =\left.-\frac {\log^2x}{2x^2}\right|_1^2\left.-\frac {\log^{\phantom{2}} x}{2x^2}\right|_1^2+\frac 12\int\limits_1^2dx\,\frac 1{x^3}\end{align*}$$Can you complete the rest?
