How do you integrate #int x^3*sqrt(4+x^2) # using integration by parts? Calculus Techniques of Integration Integration by Parts 1 Answer Tom Jan 19, 2016 not by using integration by part #u = x^2+4# #du = 2x# #1/2int2x^3*sqrt(x^2+4) du# #1/2intx^2*sqrt(u)du# #x^2 = u-4# #1/2int(u-4)*sqrt(u)du# expand #1/2intu*sqrt(u)-4sqrt(u)du# #1/2intu^(3/2)-4sqrt(u)du# #1/2[2/5u^(5/2)-8/3u^(3/2)]+C# #[1/5u^(5/2)-4/3u^(3/2)]+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 9940 views around the world You can reuse this answer Creative Commons License