What is the antiderivative of $1/(x^2+4)$ ?

Question

6801 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-27T10:57:27

$int dx/(x^2+4) = 1/2arctan(x/2)+C$

Calculate the indefinite integral:

$int dx/(x^2+4)$

by substituting:

$x= 2u$

$dx = 2du$

so:

$int dx/(x^2+4) = 2 int (du)/((2u)^2+4)$

$int dx/(x^2+4) = 2int (du)/(4u^2+4)$

$int dx/(x^2+4) = 1/2 int (du)/(u^2+1)$

$int dx/(x^2+4) = 1/2arctan(u)+C$

and undoing the substitution:

$int dx/(x^2+4) = 1/2arctan(x/2)+C$

Answer 2 · 2018-03-27T10:59:28

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

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

Now let $2u=x$ and $2du=dx$

$1/4int1/((x/2)^2+1)dx=1/2int1/(u^2+1)du=1/2arctanu+"c"=1/2arctan(x/2)+"c"$

What is the antiderivative of 1/(x^2+4)1/(x^2+4)?