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

1 Answer

Sep 21, 2016

#1/2ln(x^2+1)+C, or, ln (x^2+1)^(1/2)+C, or, lnsqrt(x^2+1)+C #

Explanation:

Let #I=intx/(x^2+1)dx#.

Subst. #(x^2+1)=t rArr 2xdx=dt rArr xdx=dt/2#.

#:. I=int1/t*dt/2=1/2int1/tdt=1/2ln|t|#

#:. I=1/2ln(x^2+1)+C, or, ln (x^2+1)^(1/2)+C, or, lnsqrt(x^2+1)+C#.

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