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

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

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

$= 1/2 int dpsi qquad$

$= 1/2 psi + C$

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

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