How can I integrate $\int_{2}^{\infty}\frac{\ln x}{x^3}\,\mathrm dx$ At first I tried a $u$-substitution with $u=\ln(x)$ and $\mathrm du=\frac{1}{x}\mathrm dx$ but this didn't seem right to me.  
I just can't seem to see it for some reason.  Any hints?  
$$\int_{2}^{\infty}\frac{\ln x}{x^3}\,\mathrm dx$$
 A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
$$
\int_{2}^{\infty}{\ln\pars{x} \over x^{3}}\,\dd x
=
\left.-\,{\ln\pars{x} \over 2x^{2}}\right\vert_{2}^{\infty}
-
\int_{2}^{\infty}{1 \over -2x^{2}}\,{1 \over x}\,\dd x
=
{1 \over 8}\,\ln\pars{2} + \pars{\left.{1 \over -4x^{2}}\right\vert_{2}^{\infty}}
$$
$$\color{#0000ff}{\large%
\int_{2}^{\infty}{\ln\pars{x} \over x^{3}}\,\dd x
=
{1 \over 8}\,\ln\pars{2} + {1 \over 16}}
$$
Another method $\pars{~\mbox{with}\ n < 2~}$:
$$
\int_{2}^{\infty}{x^{n} \over x^{3}}\,\dd x
=
\left.{x^{n - 2} \over n- 2}\right\vert_{2}^{\infty}
=
{2^{n - 2} \over 2 - n}\,,\quad
\int_{2}^{\infty}{x^{n}\ln\pars{x} \over x^{3}}\,\dd x
=
{2^{n - 2}\ln\pars{2} \over 2 - n} + {2^{n - 2} \over \pars{2 - n}^{2}}
$$
$$
\int_{2}^{\infty}{\ln\pars{x} \over x^{3}}\,\dd x
=
{2^{0 - 2}\ln\pars{2} \over 2 - 0} + {2^{0 - 2} \over \pars{2 - 0}^{2}}
=
{1 \over 8}\,\ln\pars{2} + {1 \over 16}
$$
