Integrate $1/(4+x^2) dx$ ?

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Answer 1 · 2018-05-31T13:08:42

$int1/(4+x^2)"d"x=1/2arctan(1/2x)+"c"$

$int1/(4+x^2)"d"x=int1/(4(1+x^2/4))"d"x=1/4int1/(1+(x/2)^2)"d"x$

Let $u=x/2$ and $"d"u=1/2"d"x$

Then

$1/4int1/(1+(x/2)^2)"d"x=1/2int1/(1+u^2)"d"u$

This is a well-known integral which evaluates to

$1/2arctanu=1/2arctan(1/2x)+"c"$

Integrate 1/(4+x^2) dx1/(4+x^2) dx ?