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

Answer 1

Evelyn Agard

#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#

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Answer 2

Scarlett Allison

The antiderivative of (\frac{2 + x^2}{1 + x^2}) is (\ln(1 + x^2) + 2x - 2\arctan(x) + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.