Integral of a polynomial fraction: $\int \frac{4x^2-3x+2}{4x^2-4x+3}dx$ I have initially used long division to obtain: $$1+\frac{x-1}{4x^2-4x+3}$$
Now, I'm having difficulty solving $$\int \frac{x-1}{4x^2-4x+3}dx$$
Even if I do the $u$-substitution i.e. $u=4x^2-4x+3$ hence, $du=8x-4$ and $dx=\frac{1}{8x-4}du$, I can't seem to figure out how to deal with the $x-1$ in the numerator.
Any help regarding this will be appreciated. Thank you.

Hints:
$$2\frac{x-1}{4x^2-4x+3}=\frac{2x-2}{(2x-1)^2+2}=\frac{2x-1}{(2x-1)^2+2}-\frac1{(2x-1)^2+2}.$$
In the first term, the numerator is the derivative of the denominator (to a coefficient). In the second, we recognize the derivative of the arc tangent, modulo simple transformations.
