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

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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"#