How do you integrate #3/(x^2+4)#?

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Answer 1 · 2015-02-20T18:44:21

I would start by factorizing #4# from the denominator:

#int3/(x^2+4)dx=int3/4*1/(1+x^2/4)dx=#

Now I set:
#x^2/4=t^2# and so #x=2t# giving:

#dx=2dt#

Substituting in the integral:

#=int3/4*1/(1+t^2)2dt=3/2arctan(t)=#

Going back to #x#:

#=3/2arctan(x/2)+c#