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 10118 views around the world You can reuse this answer Creative Commons License