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How to solve this: integrate 1/(x^2+9)dx?
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look in the back cover of your calc book for integrals of the form \[\int\frac{dx}{a^2+x^2}\] and you will see that it is \[\frac{1}{a}\tan^{-1}(\frac{x}{a})\]
you know \[\Large \frac{d}{dx}(\tan^{-1}(x)=\frac{1}{x^2+1}\]
guys mind if I watch this?
or recall that \[\frac{d}{dx}[\tan^{-1}(x)]=\frac{1}{1+x^2}\] and adjust for the constant
what @sami-21 said
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