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am I doing this the right way? $\int_{-1}^{1} \frac{2x}{x^2 - 9}dx = \int_{-1}^{1} \frac{d(x^2 - 9)}{x^2 - 9}dx = \left[\ln|x^2 - 9|\right]_{-1}^{1} = 0$

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3 Answers 3

up vote 10 down vote accepted

The calculation is correct.

You might note instead that in the interval you are integrating over, our function is well-behaved (integrable). It is an odd function, and we are integrating over an interval symmetric about $x=0$. So we can conclude that the integral is $0$ without doing any calculation at all.

Recall that a function $f(x)$ is odd if $f(-x)=-f(x)$ for all $x$.

You might want to sketch the curve, or have software do it for you. The negative "area" from $-1$ to $0$ exactly cancels the positive area from $0$ to $1$.

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I think that it is important to mention that by well-behaved you mean integrable. $1/x$ is odd alright, but we still can't say that $\int_{-1}^1 1/x=0$. –  tomasz Mar 2 '13 at 23:47
    
Thank you. I have inserted the term "integrable," but also kept "well-behaved," to try to balance intuition and correctness. –  André Nicolas Mar 3 '13 at 1:25

It's completely right, minus one small notational flaw, which might come back to bite in a totally different problem.

$$\int \frac{2x}{x^2 - 9}dx = \int \frac{d(x^2 - 9)}{x^2 - 9}$$ not $$\int \frac{d(x^2 - 9)}{x^2 - 9}dx $$

Thinking of it as $$\int \frac{2x\,dx}{x^2 - 9}$$ helps.

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Not quite. Use u-substitution and set u=x^2-9.

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(not my -1) This is actually a form of u-substitution. Some people learn it this way instead. –  anorton Mar 3 '13 at 1:22

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