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:

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

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