Evaluating the indefinite integral $\int \frac {x^7-1}{\log x}dx$
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.
You said you originally wanted to solve the definite integral from $x = 0$ to $1$, 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 \quad (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}
Integrating $I'$
\begin{align} I(\alpha) &= \int \frac{1}{\alpha + 1} d \alpha \\ &= \ln(\alpha + 1) + C \quad (2) \end{align}
Note that in $(1)$ we have $I(0) = 0$. Applying this to $(2)$, we find $C = 0$, so
\begin{align} I(\alpha) &= \ln(\alpha + 1) \\ \end{align}
Evaluating at $\alpha = 7$ gives the result.