How do you integrate #int(x)/((x+4)(x+6)(x+1))# using partial fractions? Calculus Techniques of Integration Integral by Partial Fractions 1 Answer Leland Adriano Alejandro Jan 15, 2016 #2/3 ln (x+4) -3/5 ln (x+6) -1/15 ln (x+1)+C# C=the constant of inegration Explanation: #x=A(x^2+7x+6)+B(x^2+5x+4)+C(x^2+10x+24)# #A=2/3# #B=-3/5# #C=-1/15# #int x/((x+4)(x+6)(x+1)) dx = int A/(x+4) dx+ int B/(x+6) dx+int C/(x+1) dx# Answer link Related questions How do I find the partial fraction decomposition of #(2x)/((x+3)(3x+1))# ? How do I find the partial fraction decomposition of #(1)/(x^3+2x^2+x# ? How do I find the partial fraction decomposition of #(x^4+1)/(x^5+4x^3)# ? How do I find the partial fraction decomposition of #(x^4)/(x^4-1)# ? How do I find the partial fraction decomposition of #(t^4+t^2+1)/((t^2+1)(t^2+4)^2)# ? How do I find the integral #intt^2/(t+4)dt# ? How do I find the integral #int(x-9)/((x+5)(x-2))dx# ? How do I find the integral #int1/((w-4)(w+1))dw# ? How do I find the integral #intdx/(x^2(x-1)^2)# ? How do I find the integral #int(x^3+4)/(x^2+4)dx# ? See all questions in Integral by Partial Fractions Impact of this question 1104 views around the world You can reuse this answer Creative Commons License