How do you integrate #int x^3/[x(x^2+2x+1)]# using partial fractions? Calculus Techniques of Integration Integral by Partial Fractions 1 Answer Ananda Dasgupta Apr 5, 2018 # x-2ln|x+1|-1/(x+1)+C# Explanation: # x^3/[x(x^2+2x+1)] = x^2/(x+1)^2 =(1-1/(x+1))^2# #qquad = 1-2/(x+1)+1/(x+1)^2# Thus #int x^3/[x(x^2+2x+1)]dx = int ( 1-2/(x+1)+1/(x+1)^2)dx# #qquad = x-2ln|x+1|-1/(x+1)+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 1145 views around the world You can reuse this answer Creative Commons License