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

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Answer 1 · 2018-05-10T00:30:56

#x+tan^-1(x)+C#

Given: #int(2+x^2)/(1+x^2) \ dx#.

We can split as follows:

#=int(1+1+x^2)/(1+x^2) \ dx#

#=int(1+x^2)/(1+x^2)+1/(1+x^2) \ dx#

#=int(1+x^2)/(1+x^2) \ dx+int1/(1+x^2) \ dx#

The first integral is trivial, and the second one is a common integral which comes out to #arctan(x)# or #tan^-1(x)#.

#=int1 \ dx+tan^-1(x)+C#

#=x+tan^-1(x)+C#