Not sure what I did wrong in this integral

Jan 2013
82
1
Montreal
Hi,

The integral is \(\displaystyle \int \frac{x-1}{x^2-4x-5}\).

I tried to first do a little manipulation and add \(\displaystyle -1 + 1\) to the integral to get:

\(\displaystyle \int \frac{x-2}{x^2-4x-5} + \int \frac{1}{x^2-4x-5}\)

The first one solves easily using u sub, the second I solved using partial fractions:

\(\displaystyle \frac{A}{x-5} + \frac{B}{x+1}\)

The coefficients solved to \(\displaystyle A = \frac{1}{6}\) and \(\displaystyle B = -\frac{1}{6}\)

That should give me a final answer of \(\displaystyle \frac{1}{2}ln|{x^2-4x-5}| + \frac{1}{6} ln|{x-5}| - \frac{1}{6}ln|{x+1}|\)

The problem is that the integral is supposed to be a definite integral from 0 to 4, and the function I got as an answer isn't continuous over that interval. If I factor out the coefficients and combine the ln's using log properties I get something that is solvable but that doesn't give the right answer.

Can anyone see what I did wrong?

Thanks!
 
Jun 2008
1,389
513
Illinois
The problem is that \(\displaystyle \int \frac {dx} {x-5} = \ln(x-5) \) only in the domain x>5. For the interval 0 to 4 you must make a substitution:

Let \(\displaystyle u = -(x-5)\). Then \(\displaystyle \int_0 ^4 \frac {dx} {x-5} = \int_5 ^1 \frac {du} u = \ln(1)-\ln(5) = -\ln(5)\)
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
I graphed your antiderivative on [0,4] ... looks continuous to me.
 

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Jun 2008
1,389
513
Illinois
skeeter: I think you may have interpreted the OP's "|" marks as absolute values - they actually should be parentheses. His problem is that ln(x-5) is not defined for x<=5.
 
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skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
I disagree ... the antiderivative requires the absolute value to be correct.

The integrand is continuous over [0,4] and value of the definite integral is

\(\displaystyle \int_0^4 \frac{x-1}{x^2-4x-5} \, dx = -\frac{\ln(5)}{3} \approx -0.536479...\)

his antiderivative was \(\displaystyle \frac{1}{2}\ln|x^2-4x-5| + \frac{1}{6}\ln|x-5| - \frac{1}{6}\ln|x+1|\)

the absolute value is required for the first two terms, not for the last.

evaluating this antiderivative from x = 0 to x = 4 yields ...

\(\displaystyle \left[\frac{1}{2}\ln(5) + 0 - \frac{1}{6}\ln(5)\right]-\left[\frac{1}{2}\ln(5) + \frac{1}{6}\ln(5) - 0\right] = -\frac{\ln(5)}{3}\)


rational functions that yield log functions when integrated can be evaluated when the log's argument is negative, hence the need for absolute value in the antiderivative. as a simple example ...

\(\displaystyle \int_{-6}^{-1} \frac{1}{x} \, dx = \left[\ln|x|\right]_{-6}^{-1} = \ln|-1| - \ln|-6| = -\ln(6)\)


the attached image is a calculator evaluation of the original integral.
 

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Jun 2008
1,389
513
Illinois
OK, thanks skeeter. I assumed that the OP's issue was that he wasn't using the absolute values, even though he included them in his answer. So, you're right - it seems there is really no issue here.