How do I compute this improper integral?

Question

$$\int_{0}^{1}\dfrac{1}{2x^2-x}dx$$

This is a Type II improper integral because the function is undefined at $x=\frac{1}{2}$. If I were to do this problem on my own I would split this integral in two,

$$\int_{0}^{\frac{1}{2}}\dfrac{1}{2x^2-x}dx+\int_{\frac{1}{2}}^{1}\dfrac{1}{2x^2-x}dx$$

now I have two more improper integrals. I am unsure how to split these two integrals further so I would be able to compute them. Could someone offer me a hint?

Answer 1

$$\frac1{2x^2-x}=\frac2{2x-1}-\frac1x$$

Added: If you’re asking about splitting one of the two integrals into two, to get three integrals altogether, note that the original integral is improper at $0$ as well as at $\frac12$: you need to break it in two so as to get integrals that are improper at only one endpoint.

Answer 2

An integral with $f=O(x^{-2})$ from $0$ (or wherever $x^{-2}$ creates a pole) should diverge.