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

1 Answer
Monzur R.
May 31, 2018

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

Explanation:

#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"#

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