Find $\int_0^1 \frac{x-1}{\ln (x)} \, dx$

I would like to point out that I've already read the article located at Closed-form Expression for Definite Integral of (x-1)/ln(x) dx over a non-negative domain

I am interested in finishing this problem and finding the closed form of this integral using Leibniz's Rule. So first:

Leibniz's Rule: Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be continuous on the rectangle $R= [a,b] \times [c,d]$ and let $$F(y) = \int_a^b f(x,y) \, dx$$ for each $y \in [c,d]$. If $\frac{\partial f}{\partial y}$ exists and is continuous on $R$, then $F$ is differentiable on $[c,d]$ and $$F'(y) = \int_a^b \frac{\partial f}{\partial y} f(x,y) \, dx$$

Find

$$\int_0^1 \frac{x-1}{\ln (x)} \, dx$$

Solution.

To find this antiderivative, first we consider the integral

$$F(y) = \int_0^1 \frac{x^y-1}{\ln(x)} \, dx$$

We may now use Leibniz's Rule to obtain

$$F'(y) = \int_0^1 \frac{x^y \ln(x)}{\ln(x)} \, dx = \int_0^1 x^y \, dx = \frac{1}{y+1}$$

So now we have $F$ as a function of $y$ with no $x$ appearing anywhere.

$$F'(y) = \frac{1}{y+1} \implies F(y) = \ln(y+1)+C$$

It is at this point that my progress comes to a halt. My question is: how do I find $C$? and once I find $C$, how to incorporate that into the original integral? I do also understand that the original integral is essentially the $\frac{x^y-1}{\ln(x)}$ where $y=1$.

The integral you want to find is $$F(1)=\int_0^1\frac{x-1}{\ln(x)}\,\mathrm{d}x.$$ A boundary condition that you have is that $$F(0)=\int_0^1\frac{1-1}{\ln(x)}\,\mathrm{d}x=\int_0^1{0}\,\mathrm{d}x=0.$$ Keeping this in mind, $$\forall{y\in[0,1]},\int_0^1x^y\,\mathrm{d}x=F'(y)=\frac1{y+1}.$$ Therefore, $$\int_0^1F'(y)\,\mathrm{d}y=\int_0^1\frac1{y+1}\,\mathrm{d}y.$$ The left-hand side of the above equation is $$\int_0^1F'(y)\,\mathrm{d}y=F(1)-F(0)=F(1)$$ while the right-hand side is $$\int_0^1\frac1{y+1}\,\mathrm{d}y=\ln(1+1)-\ln(1+0)=\ln(2).$$ Therefore, $$F(1)=\ln(2).$$