What is the antiderivative of # (x^2-1) / (x^2+1)#? Calculus Introduction to Integration Integrals of Rational Functions 1 Answer Roy E. Dec 20, 2016 #x-2 tan^-1 x+C# Explanation: #int(x^2-1)/(x^2+1)dx# #=int(x^2+1-2)/(x^2+1)dx# #=int1-2/(1+x^2)dx# #=x-2 tan^-1x+C#. (If the quoting standard integral #int1/(1+x^2)dx=tan^-1x# is not acceptable, substitute #x=tan u#, #dx=sec^2 u du# and use #sec^2u=1+tan^2u# to reduce the integral to #int 1du#.) Answer link Related questions How do you integrate #(x+1)/(x^2+2x+1)#? How do you integrate #x/(1+x^4)#? How do you integrate #dx / (2sqrt(x) + 2x#? What is the integration of #1/x#? How do you integrate #(1+x)/(1-x)#? How do you integrate #(2x^3-3x^2+x+1)/(-2x+1)#? How do you find integral of #((secxtanx)/(secx-1))dx#? How do you integrate #(6x^5 -2x^4 + 3x^3 + x^2 - x-2)/x^3#? How do you integrate #((4x^2-1)^2)/x^3dx #? How do you integrate #(x+3) / sqrt(x) dx#? See all questions in Integrals of Rational Functions Impact of this question 1187 views around the world You can reuse this answer Creative Commons License