Evaluate the integral $\int_{0}^1\frac{x-6}{x^2-6x+8}dx$ I used partial fractions to get:
$\displaystyle\frac{A}{x-4}+\frac{B}{x-2}= \frac{x-6}{(x-4)(x-2)}$
$A = -1$
$B = 2$
$\displaystyle\int_{0}^1\frac{-1}{x-4}+\frac{2}{x-2}dx$
Found the anti-derivative to be:
$(-\ln|x-4| + 2 \ln|x-2|)_0^{1}$
My answer came out to be around -1.1, what am I doing wrong? Thanks for any help.
Edit- Apparently WebAssign thinks that I'm wrong for some reason:


In the interval $[0,1]$ we have $\frac{-5}{3}\leq\frac{x-6}{x^2-6x+8}\leq\frac{-3}{4}$, hence:
$$ \frac{-5}{3}\leq\int_0^1\frac{x-6}{x^2-6x+8}\leq\frac{-3}{4} $$
