How do you find the indefinite integral of #int (x^2-6x-20)/(x+5)#? Calculus Introduction to Integration Integrals of Rational Functions 1 Answer Andrea S. Nov 30, 2016 #int frac (x^2-6x-20) (x+5) dx = (x+5)^2/2 -16(x+5) +35 ln (x+5)# Explanation: Substitute #x=t-5# #int frac (x^2-6x-20) (x+5) dx = int (dt)/t((t-5)^2 -6(t-5) -20) =# #= int (dt)/t(t^2-10t +25 -6t+30 -20) =# #= int (dt)/t(t^2-16t +35) = int t dt -16 int dt +35 int (dt)/t = # # = (t^2)/2 -16t +35 ln(t) = (x+5)^2/2 -16(x+5) +35 ln (x+5)# 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 1190 views around the world You can reuse this answer Creative Commons License