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How do you find the integral of #(1+x) / (1+x^2)#?

1 Answer

Sasha P. · Truong-Son N.

Sep 22, 2015

#arctanx+1/2ln(1+x^2)+C#

Explanation:

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

Let:
#t = 1+x^2#
#dt = 2xdx => xdx=dt/2#

#I = arctanx+1/2int 1/t dt#

#= color(blue)(arctanx+1/2ln(1+x^2)+C)#

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