It's the integral from 1 to 10 of x divided by x^2-4. I'm assuming u would be x^2-4 making du 2x so du/2 would be x. Then would I just be left with 1/2 the integral of 1/u and would have to solve it from there?

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- Feb 17th 2009, 03:36 PMfattydqEvaluating this definite integral
It's the integral from 1 to 10 of x divided by x^2-4. I'm assuming u would be x^2-4 making du 2x so du/2 would be x. Then would I just be left with 1/2 the integral of 1/u and would have to solve it from there?

- Feb 17th 2009, 03:38 PMJester
- Feb 17th 2009, 03:41 PMfattydq
- Feb 17th 2009, 05:33 PMKrizalid
It must be a typo on the problem or those aren't the right bounds.

You're learning how to evaluate definite integrals, but you can't do this one because the function ain't continuous at $\displaystyle x=2.$ Regarding those bounds, we have $\displaystyle 1\le x\le10$ and $\displaystyle 2$ belongs there.

If the question does not include the $\displaystyle -2$ neither $\displaystyle 2,$ then you can evaluate the integral.