Evaluate $\int\sqrt[5]{\frac{x+5}{x-5}}\,\mathrm dx$ How to compute this integral? I stuck at a point where I get $\displaystyle\int\frac{1}{t^5-1}+ \cdots
$ $$\int\sqrt[5]{\frac{x+5}{x-5}}\,\mathrm dx$$ using $\displaystyle t=\sqrt[5]{\frac{x+5}{x-5}}$
 A: Substituting $t=\sqrt[\Large5]{\frac{x+5}{x-5}}$,
$$
\begin{align}
\int\sqrt[\Large5]{\frac{x+5}{x-5}}\,\mathrm{d}x
&=\int t\,\mathrm{d}\frac{10}{t^5-1}\\
&=\frac{10t}{t^5-1}-10\int\frac{\mathrm{d}t}{t^5-1}\\
\end{align}
$$
Substituting $\sin(2\pi/5)u=t-\cos(2\pi/5)$ and $\sin(4\pi/5)v=t-\cos(4\pi/5)$,
$$
\begin{align}
&\frac1{t^5-1}\\
&=\frac15\sum_{k=0}^4\frac{e^{2\pi ik/5}}{t-e^{2\pi ik/5}}\\
&=\frac{2t\cos(2\pi/5)-2}{t^2-2t\cos(2\pi/5)+1}+\frac{2t\cos(4\pi/5)-2}{t^2-2t\cos(4\pi/5)+1}+\frac1{t-1}\\
&=2\frac{\cos(2\pi/5)(t-\cos(2\pi/5))-\sin^2(2\pi/5)}{(t-\cos(2\pi/5))^2+\sin^2(2\pi/5)}\\
&+2\frac{\cos(4\pi/5)(t-\cos(4\pi/5))-\sin^2(4\pi/5)}{(t-\cos(4\pi/5))^2+\sin^2(4\pi/5)}\\
&+\frac1{t-1}\\[6pt]
&=2\frac{\cot(2\pi/5)u-1}{u^2+1}+2\frac{\cot(4\pi/5)v-1}{v^2+1}+\frac1{t-1}
\end{align}
$$
so that
$$
\begin{align}
\int\frac1{t^5-1}\,\mathrm{d}t
&=\cos(2\pi/5)\log(u^2+1)-2\sin(2\pi/5)\arctan(u)\\
&+\cos(4\pi/5)\log(v^2+1)-2\sin(4\pi/5)\arctan(v)\\[6pt]
&+\log(t-1)\\[6pt]
&+C
\end{align}
$$
Therefore, using the substitutions above, which are unwieldy to write, but simple to compute,
$$
\begin{align}
\int\sqrt[\Large5]{\frac{x+5}{x-5}}\,\mathrm{d}x
&=\frac{10t}{t^5-1}\\[3pt]
&-10\cos(2\pi/5)\log(u^2+1)+20\sin(2\pi/5)\arctan(u)\\[6pt]
&-10\cos(4\pi/5)\log(v^2+1)+20\sin(4\pi/5)\arctan(v)\\[6pt]
&-10\log(t-1)\\[6pt]
&-10\,C
\end{align}
$$
A: setting $$t=\sqrt[5]\frac{x+5}{x-5}$$ we get $$x=\frac{5(t^5+1)}{t^5-1}$$ and we get
$$dx=-\frac{50t^4}{(t-1)^2(t^4+t^3+t^2+t+1)^2}dt$$ thus our integral is
$$-50\int\frac{t^5}{(t-1)^2(t^4+t^3+t^2+t+1)^2}dt$$ this is not so easy to solve but rational and can be solved explicitely
A: You are looking at Gauss Hypergeometric ${}_2F_1$. See here
or
here.
Compare e.g.,
$$\int \sqrt[5]{\frac{x+5}{x−5}}\,dx = \int{(x+1)^{\frac{1}{5}}(x-8+1)^{-\frac{1}{5}}}dx$$
with http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/07/01/01/.
I would not expect to find a "simpler" answer than one expressed in terms of ${}_2F_1$.
