How do you integrate #int x^2 e^(-x) dx # using integration by parts? Calculus Techniques of Integration Integration by Parts 1 Answer Lovecraft Jan 9, 2016 #I = -e^(-x)(x^2 + 2x + 2) + c# Explanation: #I = intx^2e^(-x)dx# Say #u = x^2# so #du = 2x# and #dv = e^(-x)# so #v = -e^(-x)# #I = -x^2e^(-x) + 2intxe^(-x)dx# Say #u = x# so #du = 1# and #dv = e^(-x)# so #v = -e^(-x)# #I = -x^2e^(-x) + 2(-xe^(-x) + inte^(-x)dx)# #I = -x^2e^(-x) -2xe^(-x) + 2inte^(-x)dx# #I = -x^2e^(-x)-2xe^(-x)-2e^(-x)+c# #I = -e^(-x)(x^2 + 2x + 2) + c# Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 1483 views around the world You can reuse this answer Creative Commons License