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

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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.!

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