How do I solve $\int_1^{2}\frac{(\ln{x})^2}{x^3} dx$?

Question

$\int_1^{2}\frac{(\ln{x})^2}{x^3} dx$

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?

Answer 1

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 $dv=e^{-2t}dt$

Answer 2

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?

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}