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This might be a stupid question but what is the integral of $\frac{1}{1 + 9x^2}$? I want to think it's $\tan^{-1}(3x) + c$ but the book I'm working in says $\frac{1}{3}\tan^{-1}(3x) + c$.

How did they get $1/3$?

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    Note that you integrate anything at wolframalpha (google it), then click show steps, and it will give you the step by step instructions for integrating!2011-03-17
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    @Fdart: This is not particularly good advice. You didn't even answer --or even address in any way-- OP's question, which has a very straightforward answer if you just think about it rather than re-route yourself automatically to a computation tools...2011-03-17
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    You can always check to see if your answer is correct by differentiation. If you differentiate $\tan^{-1}(3x)+C$, you'll get$$\frac{d}{dx}\left(\tan^{-1}(3x)+\right) = \left(\frac{1}{1+(3x)^2}\right)(3x)' = \frac{3}{1+9x^2}.$$So your answer is off by a factor of $3$.2011-03-17
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    @Fdart: in this case show steps in Alpha yields nothing. You get the integral, but no steps. But it is a reasonable thing to try first, and for OP to indicate that s/he OP has tried it and doesn't understand.2011-03-17

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