# How do you integrate #1/(x^2 + 9)#?

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mason m
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Jun 24, 2016

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We will try to put this in the form of the arctangent integral:

#int1/(u^2+1)du=arctan(u)+C#

So here, we see that:

#int1/(x^2+9)dx=int1/(9(x^2/9+1))dx=1/9int1/((x/3)^2+1)dx#

Let

#=1/3int(1/3)/((x/3)^2+1)dx=1/3int1/(u^2+1)du=1/3arctan(x/3)+C#

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mason m
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Jun 20, 2017

Let

Since

Reversing the original substitution

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