How do you integrate #int x^3*sqrt(4+x^2) # using integration by parts?

Question

9977 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-19T10:03:57

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#