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! 

Hint: $\dfrac{d}{dx}(\log_e(x)) = \dfrac 1x$.
Putting $u = \log_e x \implies du = \frac 1x\,dx$. 
That gives you:
$$ \int \frac{dx}{x(\log_e x)^{7/8}} = \int (\underbrace{\log_e x}_{u})^{-7/8}\cdot \underbrace{\left(\frac 1x \,dx \right)}_{du}=\int u^{-7/8} \,du$$
Can you take it from here (using the power rule)?
SPOILER (Scroll over to check your work!)

$$\int u^{-7/8}\,du = \frac{u^{1/8}}{1/8} + C = 8u^{1/8} + C = 8(\log_e x)^{1/8} + C$$

