# 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:

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Why do you think you have done something wrong? The integrand is negative, so you should expect the integral to be negative. – Gerry Myerson Mar 4 '12 at 22:26
Exact result is $\ln\frac{1}{3}\approx -1.09$. So your computations are correct. – Norbert Mar 4 '12 at 22:27
WebAssign may be looking for the answer in a particular format, such as $-\log3$, or it may be looking for a certain number of decimals of accuracy. What happens when you click on "Read it"? Does that explain what WebAssign is looking for? – Gerry Myerson Mar 4 '12 at 22:53
@Gerry that did the trick. I typed in: "-ln(3)-ln(4)+2ln(2)" – mathisnotmyforte Mar 4 '12 at 23:08
@mathisnotmyforte I find it really disappointing that you choose/prefer $-1.1$ as an answer rather than $-\log 3$. – Pedro Tamaroff Mar 6 '12 at 1:17

You're doing nothing wrong.

Simply substitute the boundary numbers, to get

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

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

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

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

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

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