I have a question that I have no idea how to start:
$$\int_{0}^{1} \frac{x-1}{\ln x}dx$$
So in one of my classes, I learned a technique called parametric integration. But I have no idea how to use the technique with this question.
Any hints/help would be great!
The parametric integrand you want is as follows. Put
$$f(p)=\int_0^1\frac{x^p-1}{\ln x}dx$$
Then
$$f'(p) =\int_0^1\frac{\partial}{\partial p}\left(\frac{x^p-1}{\ln x}\right)dx=\int_0^1 x^p\;dx=\left.\frac{x^{p+1}}{p+1}\right|_0^1=\dfrac{1}{p+1}$$
so $f(p)= \ln(p+1)+ C$, where $C$ is to be determined. To find $C$, just note that $f(0)=0$ since the integrand vanishes when $p=0$; this means $C=0$, so $$f(p)=\ln(p+1).$$
Note: In your specific case, take $p=1$ to get $\int_0^1\frac{x-1}{\ln x}dx=\boxed{\ln 2}.$