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

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Answer 1 · 2016-07-04T08:09:00

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

#int dx qquad 1/(x^2+4)#

use sub #x^2 = 4 tan^2 psi#

so #x = 2 tan psi, dx = 2 sec^2 psi \ d psi#

integral becomes

#int dpsi qquad 2 sec^2 psi *1/(4 tan^2 psi^2+4)#

#= 1/2 int dpsi qquad (sec^2 psi)/(sec^2 psi)#

#= 1/2 int dpsi qquad #

#= 1/2 psi + C#

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