Integrate $\int\frac{5x-7}{x^2-3x+2}$

Question

I want to integrate $\int\frac{5x-7}{x^2-3x+2}$ but my result differs from the one on Wolframalpha http://www.wolframalpha.com/input/?i=integrate+%285x-7%29%2F%28x%5E2-3x%2B2%29

I did the following steps:

$$\frac{5x-7}{(x-2)(x-1)} = \frac{A}{x-2}+\frac{B}{x-1}$$ $$5x-7 = A(x-1)+B(x-2)$$ $$5 = A + B$$ $$-7 = -A-2B$$ $$A=5-B\\ -7= -(5-B)-2B\\ -2 = -B\\ 2 = B$$

Therefore $A = 3$ and $B=2$

$$\int\frac{3}{x-2}+\int\frac{2}{x-1}=3\ln(x-2)+2\ln(x-1) + C$$

While on Wolframalpha it is $3\ln(2-x)+2\ln(1-x)$ Where did I do the error?

Answer 1

Neither is quite right. We have $\int \frac{1}{u}\,du=\ln(|u|)+C$.

Answer 2

They are both valid. $$\frac{d}{dx}\ln(a-x)=\frac{-1}{a-x}=\frac{1}{x-a}=\frac{d}{dx}\ln(x-a)$$