How do you integrate #1/(x^2+4)#?

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Answer 1 · 2016-06-24T13:29:56

#1/2arctan(x/2)+C#

Our goal should be to make this mirror the arctangent integral:

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

To get the #1# in the denominator, start by factoring:

#int1/(x^2+4)dx=int1/(4(x^2/4+1))dx=1/4int1/(x^2/4+1)dx#

Note that we want #u^2=x^2/4#, so we let #u=x/2#, which implies that #du=1/2dx#.

#1/4int1/(x^2/4+1)dx=1/2int(1/2)/((x/2)^2+1)dx=1/2int1/(u^2+1)du#

This is the arctangent integral:

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