How can I prove the integral? Prove that 
$$
\int\frac{dx}{x(\log_e x)^{7/8}} = 8(\log_e x)^{1/8}
$$
I am totally lost on this subject. Any help how to prove this is appreciated! 

Firstly lets go with the convention of $ \log_ex=\ln x $, now $(\ln x)'=\frac {1}{x}$ so we have:
$\int (\frac {1}{x}(\ln x)^{-7/8})dx=\int(\ln x)^{-7/8})d(\ln x)=8(\ln x)^{1/8}+C$
