<s>[INST] Closed form of integral $\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} \, dx$ Originally this integral was interesting for me as it represents the specific harmonic number $H_{-\frac{1}{5}}$ and also because Mathematica returned the wrong value $0$ for the integral. I posted the probem here https://mathematica.stackexchange.com/questions/215089/possible-bug-integrate1-x-1-5-1-x-x-0-1-0
Later I found a way to get a reasonable result from Mathematica which, finally I could simplify to this form.
$$\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} \, dx=\frac{\pi}{2}   \sqrt{1+\frac{2 \sqrt{5}}{5}}-\frac{5}{4}  \log (5)-\frac{\sqrt{5}}{2}  \log \left(\frac{1}{2} \left(\sqrt{5}+1\right)\right)$$
So a presumed bug by has turned finally into a nice statement. Can you prove it?
 [/INST] Note
$$I=\int_0^1 \frac{1-\frac{1}{\sqrt[5]{x}}}{1-x} dx
\overset{ x=t^5}=\int_0^1 \frac{-5t^3}{t^4+t^3+t^2+t+1}dt$$
$$=\sqrt5\int_0^1\left( \frac{2\phi_-t+1}{t^2+2\phi_-t+1}
- \frac{2\phi_+t+1}{t^2+2\phi_+t+1}\right)dt = \sqrt5 J(\phi_-) - \sqrt5 J(\phi_+)$$
where $\phi_{+}=\frac{1+\sqrt5}4= \cos \frac\pi5$, $\phi_{-}=\frac{1-\sqrt5}4= \cos \frac{3\pi}5$ and
$$J(\phi) = \int_0^1 \frac{2\phi t+1}{t^2+2\phi t+1}dt
=  \phi \ln( t^2+2\phi t+1 )|_0^1 
+ \frac{1-2\phi^2}{\sqrt{1-\phi^2}} \tan^{-1}\frac{t+\phi}{\sqrt{1-\phi^2}}\bigg|_0^1$$
Then
\begin{align} 
I= &\sqrt5 [\phi_-\ln( 2+2\phi_-) - \phi_+\ln( 2+2\phi_+)]
 + \frac{\sqrt5\pi}{10} \left( 3\frac{1-2\phi_-^2}{\sqrt{1-\phi_-^2}}  - \frac{1-2\phi_+^2}{\sqrt{1-\phi_+^2}} \right)\\
 = &\frac{\sqrt5-5}4\ln\frac{\sqrt5-1}{2\sqrt5} - \frac{\sqrt5+5}4\ln\frac{\sqrt5+1}{2\sqrt5}
 + \frac{\sqrt5\pi}{10} \left( \frac{3(1+\sqrt5)}{\sqrt{10+2\sqrt5}} - \frac{1-\sqrt5}{\sqrt{10-2\sqrt5}} \right) \\= &-\frac{\sqrt5}2\ln\frac{\sqrt5+1}2 -\frac{5}4\ln5
 + \frac{\sqrt5\pi}{10}\sqrt{5+2\sqrt5}\\
=&\frac{\pi}{2}   \sqrt{1+\frac{2 \sqrt{5}}{5}}-\frac{5}{4}  \ln5-\frac{\sqrt{5}}{2}  \ln \frac{\sqrt{5}+1}2
\end{align}
</s>