TITLE: Evaluating the indefinite integral $\int \frac {x^7-1}{\log x}dx$
QUESTION [6 upvotes]: The integral to be evaluated is  $$\int \frac {x^7-1}{\log x}dx$$
The actual question was to find the definite integral of the above in the limits $0$ to $1$. I could not apply any of the usual tricks to find that integral, hence I thought finding the indefinite integral first and then applying the limits would be the best way to evaluate the same.
I tried applying trigonometric substitution, which did'nt work. I also tried putting $x=e^t$, hoping to simplify the integral but that did not give me any results either. 
How should I proceed to evaluate this integral?

REPLY [6 votes]: You said you originally wanted to solve the definite integral for $x \in [0, 1]$ so I will demonstrate how to do it without resorting to finding the primitive. To solve this integral, we use the Feynman trick of differentiating under the integral. First, consider
\begin{align}
I(\alpha) &= \int_{0}^{1} \frac{x^{\alpha} - 1}{\log (x)} dx \tag 1
\end{align}
Differentiating with respect to $\alpha$ gives
\begin{align}
I'(\alpha) &= \int_{0}^{1} \frac{d}{d \alpha} \left( \frac{x^{\alpha} - 1}{\log (x)} \right) dx \\
&=\int_{0}^{1} \frac{\log (x) \cdot x^{\alpha}}{\log (x)} dx \\
&= \int_{0}^{1} x^{\alpha} dx \\
&= \frac{1}{\alpha + 1}
\end{align}
and then integrating $I'$ yields
\begin{align}
\implies I(\alpha) &= \int \frac{1}{\alpha + 1} d \alpha \\
&= \ln(\alpha + 1) + C
\end{align}
Note that in $(1)$ we have $I(0) = 0$ and so we find $C = 0$. Therefore
\begin{align}
I(\alpha) &= \ln(\alpha + 1) \\
\end{align}
and evaluating at $\alpha = 7$ gives the result.