Integration question

[meghiano]

Hi,
I was working through this problem and was curious how they got the answer to be negative. The problem is as follows:

Int. {((1-sqrtx)^3)/sqrtx } dx

Thanks

 

[daon]

$\displaystyle u = (1-\sqrt{x}) \implies du = (1-x^{\frac{1}{2}})'dx = -\frac{1}{2}x^\frac{-1}{2}dx$

$\displaystyle \implies -2du = \frac{dx}{\sqrt{x}}$

So,

$\displaystyle \int(1-\sqrt{x})^3\cdot \frac{dx}{\sqrt{x}} = \int u^3(-2du)$

 

[BigGlenntheHeavy]

$\displaystyle \int \frac{(1-x^{1/2})^{3}}{x^{1/2}}dx \ Now \ let \ u \ = \ x^{1/2}, \ u^{2} \ = \ x, \ and \ 2udu \ = \ dx.$

$\displaystyle Ergo, \ we \ have \ \int \frac{(1-u)^{3}2u}{u}du \ = \ 2\int(1-u)^{3}du$

$\displaystyle Expanding \ the \ binomial, \ we \ get \ 2\int(-u^{3}+3u^{2}-3u+1)du$

$\displaystyle which \ equals \ 2[\frac{-u^{4}}{4}+u^{3}-\frac{3u^{2}}{2}+u]+ \ C = \ \frac{-u^{4}}{2}+2u^{3}-3u^{2}+2u \ +C.$

$\displaystyle Hence, \ resubstituting, \ we \ get \ -\frac{x^{2}}{2}+2x^{3/2}-3x+2x^{1/2} + \ C \ or \ -[\frac{x^{2}}{2}-2x^{3/2}+3x-2x^{1/2}] \ +C.$

Addendum: x must be greater than zero.

 

[meghiano]

Thanks, i was forgetting the prime! have a nice day