How do you find the integral of #(x^2)(e^(x-1))#? Calculus Techniques of Integration Integration by Parts 1 Answer GiĆ³ Jun 30, 2015 I found #e^(x-1)[x^2-2x+2]+c# Explanation: I used Integration by Parts (twice): 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 4372 views around the world You can reuse this answer Creative Commons License