How do you find the integral of #xe^(x/2)dx #? Calculus Techniques of Integration Integration by Parts 1 Answer Guilherme N. · Tom May 26, 2015 By part #u = x# #du = 1# #dv = e^(1/2x)# #v = 2e^(1/2x)# #=>2([xe^(1/2x)]-inte^(1/2x)dx)+C# #=>2([xe^(1/2x)]-2[e^(1/2x)])+C# #=>2e^(1/2x)(x-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 8720 views around the world You can reuse this answer Creative Commons License