<s>[INST] Solving $\int \frac{4x^2-3x+2}{4x^2-4x+3}$ with partial fractions I'm trying to understand how my textbook solved this problem but something seems a bit off.
First thing to do was to perform long division since the degree of the numerator is not less than the degree of the denominator, such a division yielded:
$$1 + \frac{x-1}{4x^2-4x-3}$$
Calculating this integral, we got to the point where it seemed plausible to do a substitution $u = 2x - 1$
$$x + \frac{1}{2} \int\frac{\frac{1}{2}(u + 1) -1}{u^2+2}du$$
Now, here's the tricky part, somehow the next simplification was:
$$x + \frac{1}{4} \int\frac{(u -1)}{u^2+2}du$$
My question is how did he do this?
 [/INST] HINT
We have that
$$\frac{4x^2-3x+2}{4x^2-4x+3}=\frac{4x^2-4x+3+x-1}{4x^2-4x+3}=1+\frac{x-1}{4x^2-4x+3}=$$
$$=1+\frac18\frac{8x-4-4}{4x^2-4x+3}=1+\frac18\frac{8x-4}{4x^2-4x+3}-\frac12\frac{1}{4x^2-4x+3}$$
$$=1+\frac18\frac{8x-4-4}{4x^2-4x+3}=1+\frac18\frac{8x-4}{4x^2-4x+3}-\frac12\frac{1}{2+(2x-1)^2}$$
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