How do you find the integral of $1/(x^2+4)$ ?

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Answer 1 · 2015-10-18T17:53:20

$1/2tan^(-1)(x/2)+C$

Recall the rule : $int1/(x^2+a^2)dx=1/atan^(-1)(x/a)+C$

We may use this rule to integrate the given function as follows :

$int1/(x^2+4)dx=int1/(x^2+2^2)dx$

$=1/2tan^(-1)(x/2)+C$

How do you find the integral of 1/(x^2+4) 1/(x^2+4) ?