How do you find the integral of #x^2 / sqrt (x-1) dx#?

Question

62558 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-02T17:05:43

# 2/15*sqrt(x-1)(3x^2+4x+8)+C.#

Let, #I=intx^2/sqrt(x-1)dx#.

Subst. #x-1=t^2. :. x=t^2+1. :. dx=2tdt#.

#:. I=int(t^2+1)^2/t*2tdt#,

#=2int(t^4+2t^2+1)dt#,

#=2(t^5/5+2*t^3/3+t)#,

#=(2t)/15(3t^4+10t^2+15)#,

#=2/15*sqrt(x-1){3(x-1)^2+10(x-1)+15}#.

# rArr I=2/15*sqrt(x-1)(3x^2+4x+8)+C.#

Enjoy Maths.!