How do you integrate #(x-2)/(x-1)#? Calculus Introduction to Integration Integrals of Rational Functions 1 Answer Jim H Sep 27, 2016 #(x-2)/(x-1) = (x-1-1)/(x-1) = (x-1)/(x-1)-1/(x-1) = 1-1/(x-1)# Explanation: #int(x-2)/(x-1) dx = int(1-1/(x-1)) dx# # = x-ln abs(x-1) +C# 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 27020 views around the world You can reuse this answer Creative Commons License