<s>[INST] What is $\int \frac{x}{x^5-7} dx$? I have tried out many trigonometric substitution like $x=\sin^{\frac{2}{5}}z$. But it did not work.
 [/INST] Substitute $x=\sqrt[5]{7}u$, and you have 
$$
\begin{align}
\int\frac{\sqrt[5]{7}u}{7u^5-7}\,\sqrt[5]{7}du
&=\frac{\sqrt[5]{49}}{7}\int\frac{u}{u^5-1}\,du\\
&=\frac{\sqrt[5]{49}}{7}\int\left(\frac{A}{u-1}+\frac{Bu+C}{u^2-2hu+1}+\frac{Du+E}{u^2-2ku+1}\right)\,du\\
\end{align}
$$
where $h=\cos(2\pi/5)=\frac{\sqrt{5}-1}{4}$ and $k=\cos(4\pi/5)=\frac{-\sqrt{5}-1}{4}$. The values of $A,B,C,D,E$ can be found using linear algebra, after recombining the three terms and comparing the coefficients of powers of $u$ in the numerator to that of $\frac{u}{u^5-1}$.
Continuing:
$$
\begin{align}
&=\frac{\sqrt[5]{49}}{7}\int\left(\frac{A}{u-1}+\frac{Bu+C}{(u-h)^2+1-h^2}+\frac{Du+E}{(u-k)^2+1-k^2}\right)\,du\\
&=\frac{\sqrt[5]{49}}{7}\int\left(\frac{A}{u-1}+\frac{B(u-h)+Bh+C}{(u-h)^2+1-h^2}+\frac{D(u-k)+Dk+E}{(u-k)^2+1-k^2}\right)\,du\\
&=\frac{\sqrt[5]{49}}{7}\left(A\ln\lvert u-1\rvert+\frac{B}{2}\ln\left((u-h)^2+1-h^2\right)+\frac{Bh+C}{\sqrt{1-h^2}}\arctan\left(\frac{u-h}{\sqrt{1-h^2}}\right)\right.\\
&\phantom{{}={}}\left.+\frac{D}{2}\ln\left((u-k)^2+1-k^2\right)+\frac{Dh+E}{\sqrt{1-k^2}}\arctan\left(\frac{u-k}{\sqrt{1-k^2}}\right)\right)+\text{constant}\\
\end{align}
$$
Note $1-h^2=\sin^2(2\pi/5)$ and $1-k^2=\sin^2(4\pi/5)$, which may help some of you needed to carry this through to an explicit end. 
The "constant" may take two different values on either side of $u=1$ where the discontinuity is.
The last step would be to back-substitute from $u$ to $x$. Then the "constant" might change values at $x=\sqrt[5]{7}$.
</s>