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

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Answer 1 · 2016-06-24T03:35:54

#1/3arctan(x/3)+C#

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 #u=x/3#, implying that #du=1/3dx#:

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

Answer 2 · 2017-06-20T04:53:05

#intdx/(x^2+9)#

Let #x=3tantheta#, implying that #dx=3sec^2thetad theta#.

#=int(3sec^2thetad theta)/(9tan^2theta+9)#

Since #tan^2theta+1=sec^2theta#:

#=1/3intsec^2theta/sec^2thetad theta=1/3intd theta=1/3theta+C#

Reversing the original substitution #x=3tantheta#:

#=1/3tan^-1(x/3)+C#