How do you integrate #x^2 e^-x#? Calculus Techniques of Integration Integration by Parts 1 Answer Konstantinos Michailidis Oct 26, 2015 It is #int x^2*e^-xdx=-e^-x*x^2+int e^-x2xdx=-x^2*e^-x-2x*e^-x+2*int e^-xdx=-e^-x*(x^2+2x+2)+c# We used integration by parts #int f'(x)*g(x)dx=f(x)*g(x)-intf(x)*g'(x)dx# 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 1432 views around the world You can reuse this answer Creative Commons License