How do you integrate #int (x^3-x^2)/ (x+3)^4# using partial fractions? Calculus Techniques of Integration Integral by Partial Fractions 1 Answer Roy E. Jan 18, 2017 #ln|x+3| +10/(x+3)-33/(2(x+3)^2)+12/(x+3)^3+c# Explanation: Let #u=x+3#, #(du)/(dx)=1#. #int((u-3)^3-(u-3)^2)/u^4 du# #=int(u^3-10u^2+33u-36)/u^4du# #=int u^-1-10u^-2+33u^-3-36u^-4dx# #=ln|u|+10/u-33/(2u^2)+12/u^3+c# 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 1427 views around the world You can reuse this answer Creative Commons License