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

Answer 1

Isaiah Allen

# 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.#

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Answer 2

Emily Ammerman

To find the integral of ( \frac{x^2}{\sqrt{x-1}} , dx ), you can use the substitution method. Let ( u = \sqrt{x-1} ). Then, ( x = u^2 + 1 ) and ( dx = 2u , du ). Substitute these expressions into the integral, and you get:

[ \int \frac{(u^2 + 1)^2}{u} \cdot 2u , du ]

Simplify the expression inside the integral, expand, and integrate term by term. Then, after integration, substitute back ( u = \sqrt{x-1} ).

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Answer from HIX Tutor

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