calculus - Evaluate $int_0^1 frac{(x^2-1)}{log x} dx$

Evaluate $$\int_0^1 \frac{(x^2-1)}{\log x} dx$$

Context:

I came across a similar integral some time ago, and now I would like to know how to tackle this. Any hint is very welcome.

Answer

Let $$f(k)=\int_0^1 \frac{x^k-1}{\log x} \mathrm{d}x$$

Then $$f'(k)=\displaystyle \int_0^1 \frac{x^k \log x}{\log x} \mathrm{d}x = \int_0^1 x^k \mathrm{d}x= \frac{1}{1+k} \iff f(k)-f(0)=\log(1+k) \iff f(k) = \log(1+k)$$

Evaluating $k=2$ we get $$f(2)=\int_0^1 \frac{x^2-1}{\log x} \mathrm{d}x=\log (1+2)= \log 3$$