Evaluate $\int\frac{\sqrt {25 - x^2}}{ x^4}$ I'm pretty sure the method used is trig substitution. But I'm having trouble setting up and solving the problem. 
 A: $\newcommand{\+}{^{\dagger}}
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With $\ds{x \equiv {1 \over u}}$:
\begin{align}
\color{#00f}{\large\int{\root{25 - x^{2}} \over x^{4}}\,\dd x}
&=\int{\root{25 - 1/u^{2}} \over 1/u^{4}}\,
\pars{-\,{\dd u \over u^{2}}}
=-\int u\root{25u^{2} - 1}\,\dd u
\\[3mm]&=-\,\half\int\root{25u^{2} - 1}\,\dd\pars{u^{2}}=-\,{\pars{25u^{2} - 1}^{3/2} \over 75}
=-\,{\bracks{25\pars{1/x}^{2} - 1}^{3/2} \over 75}
\\[3mm]&=\color{#00f}{\large -\,{\root{25 - x^{2}} \over 75x^{3}}} + \pars{~\mbox{a constant}~}
\end{align}
A: Subsititute $\theta=\sin^{-1} \dfrac{x}5\implies \dfrac{d\theta}{dx}=\dfrac{1}{\sqrt{25-x^2}}$, so we get,
\begin{align}
\\\\\\&\int\frac{\sqrt{25-x^2}}{x^4}dx
\\=&\int\frac{25-x^2}{x^4}\dfrac{1}{\sqrt{25-x^2}}d\theta
\\=&\int \dfrac{25-25\sin^2 \theta}{625\sin^4\theta}d\theta
\\=&\dfrac{1}{25}\int\dfrac{\cos^2 \theta}{\sin^4\theta}d\theta
\\=&\dfrac{1}{25}\int\cot^2\theta \csc^2\theta d\theta
\end{align}
Now subsitute $u=\cot \theta$ to finish.
