Evaluate the integral $\int_{0}^1\frac{x-6}{x^2-6x+8}dx$

Question

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:

Answer 1

-1.1 is correct according to Wolfram.

http://www.wolframalpha.com/input/?i=integrate+%28x-6%29%2F%28x%5E2-6x%2B8%29+dx+from+0+to+1

Answer 2

You're doing nothing wrong.

Simply substitute the boundary numbers, to get

$$A = -\log|1-4|+2\log|1-2| + \log|-4|-2\log|-2|$$

$$A = -\log3+2\log1 + \log4-2\log2$$

$$A = -\log3 \approx-1.09861$$

Answer 3

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

$$ \frac{-5}{3}\leq\int_0^1\frac{x-6}{x^2-6x+8}\leq\frac{-3}{4} $$