Evaluating the indefinite integral $\int \frac {x^7-1}{\log x}dx$ 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?
 A: Since you asked about the indefinite integral, I will introduce a couple of identities and definitions:
$${\rm li} (x) =   \int_0^x \frac{dt}{\ln t} \tag{1}$$
$$\operatorname{li}(x)=\operatorname{Ei}(\ln{x}) \tag{2}$$
$$\operatorname{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}t\,dt. \tag{3}$$
Where ${\rm li} (x)$ is the logarithmic integral, and $\operatorname{Ei}(x)$ is the exponential integral.

Start by using the substitution you attempted, it is a good one:
$$x=e^t \iff dx=e^t ~dt$$
Hence giving:
$$\int \frac{x^7-1}{\ln{x}}~dx=\int \frac{e^{7t}-1}{t}\cdot e^t~dt=\int \frac{e^{8t}-e^t}{t}~dt=\color{green}{\int \frac{e^{8t}}{t}~dt}-\color{blue}{\int \frac{e^t}{t}~dt}$$
Substituting $u=8t$ on the $\color{green}{\text{green}}$ integral and applying $(3)$, we have:
$$\int \frac{e^{8t}}{t}~dt=\int \frac{e^u}{u}~du=\operatorname*{Ei}(u)+c_1=\operatorname*{Ei}(8\ln{x})+c_1$$
The $\color{blue}{\text{blue}}$ integral is easy, we can apply $(3)$, then apply $(2)$.
$$\int \frac{e^t}{t}~dt=\operatorname*{Ei}(t)+c_2=\operatorname*{Ei}(\ln{x})+c_2=\operatorname*{li}(x)+c_2$$
Combining the two results, we obtain the desired result.
$$\bbox[5px,border:2px solid #C0A000]{\int \frac{x^7-1}{\ln{x}}~dx=\operatorname*{Ei}(8\ln{x})-\operatorname*{li}(x)+C}$$

Edit: Just to make the answer more slick, you can separate the integrals immediately:
$$\int \frac{x^7-1}{\ln{x}}~dx=\color{#0066ff}{\int \frac{x^7}{\ln{x}}~dx}-\color{#990000}{\int \frac{1}{\ln{x}}~dx}$$
For the $\color{#0066ff}{\text{light blue}}$ integral, we can substitute $x=e^{t/8}$ to obtain:
$$\int \frac{x^7}{\ln{x}}~dx=\int \frac{e^t}{t}~dt=\operatorname*{Ei}(t)+c_1=\operatorname*{Ei}(8\ln{x})+c_1$$
And for the $\color{#990000}{\text{red}}$ integral, we can just immediately apply $(1)$:
$$\int \frac{1}{\ln{x}}~dx=\operatorname*{li}(x)+c_2$$
Combining the two results gives the same answer as the previous approach.
A: 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.
