You have $ - \frac{1}{{2\pi n}}\int\limits_0^\infty {\frac{{d\cos 2\pi nu}}{{{{\left( {x - u} \right)}^4}}}} $
What you want to do is use integration by parts. This says that
$\int\limits_a^b {f \cdot g'\left( x \right)dx} + \int\limits_a^b {f' \cdot g\left( x \right)dx} = f \cdot g\left( b \right) - f \cdot g\left( a \right)$
This can be used on your integral, setting
$\eqalign{ & \frac{1}{{{{\left( {x - u} \right)}^4}}} = f\left( u \right) \cr & g'\left( u \right)du = d\cos 2\pi nu \cr} $
Then we have that
$\int\limits_0^\infty {\frac{1}{{{{\left( {x - u} \right)}^4}}} \cdot d\cos 2\pi nu} - \int\limits_0^\infty {\frac{4}{{{{\left( {x - u} \right)}^5}}} \cdot \cos 2\pi nudu} = \mathop {\lim }\limits_{u \to \infty } \frac{{\cos 2\pi nu}}{{{{\left( {x - u} \right)}^4}}} \cdot - \frac{{\cos 0}}{{{{\left( {x - 0} \right)}^4}}}$
It is clear the RHS limit is zero, so let's multiply by $- \frac{1}{{2\pi n}}$ to get
$ - \frac{1}{{2\pi n}}\int\limits_0^\infty {\frac{{d\cos 2\pi nu}}{{{{\left( {x - u} \right)}^4}}}} = \frac{1}{{2\pi n}}\frac{1}{{{x^4}}} - \frac{4}{{2\pi n}}\int\limits_0^\infty {\frac{{\cos 2\pi nu}}{{{{\left( {x - u} \right)}^5}}} \cdot du} $
Now it is just a matter of noticing
$\cos 2\pi nu\cdot du = \frac{{d\left( {\sin 2\pi nu} \right)}}{{2\pi n}}$
which gives
$ - \frac{1}{{2\pi n}}\int\limits_0^\infty {\frac{{d\cos 2\pi nu}}{{{{\left( {x - u} \right)}^4}}}} = \frac{1}{{2\pi n}}\frac{1}{{{x^4}}} - \frac{1}{{{\pi ^2}{n^2}}}\int\limits_0^\infty {\frac{{d\sin 2\pi nu}}{{{{\left( {x - u} \right)}^5}}}} $