How do you find the definite integral of #2 / (4+x^2) dx# from #[0, 2]#?

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Answer 1 · 2018-04-08T08:11:07

#pi/4#

The integral of #a/(a^2+x^2)# with respect to #x# is #tan^(-1)(x/a)#

#therefore# the integral of#2/(4+x^2)=tan^(-1)(x/2)# where #a=2#

So, #int_0^2 2/(4+x^2) dx#

#=[tan^(-1)(x/2)]_0^2#

#=tan^(-1)1-tan^(-1)0#

#=pi/4-0#

#=pi/4#