$\displaystyle \int(\sqrt((x+1)/(x-2))

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- Jun 5th 2012, 12:09 PMCbarker1Rigorous Integration
$\displaystyle \int(\sqrt((x+1)/(x-2))

$ - Jun 5th 2012, 12:20 PMskeeterRe: Rigorous Integration
integral sqrt((x+1)/(x-2)) dx - Wolfram|Alpha

click on "show steps" ... - Jun 5th 2012, 12:22 PMCbarker1Re: Rigorous Integration
I am confusion how that could be legal in the math world.

- Jun 5th 2012, 12:33 PMPlatoRe: Rigorous Integration
- Jun 5th 2012, 12:42 PMCbarker1Re: Rigorous Integration
I have difficulty written sentences. But i know what i am talking about.

Sorry.

I rewrite that sentence to a questions.

Why the steps in Wolfram Alpha could be correct? If yes, how? - Jun 5th 2012, 05:18 PMHallsofIvyRe: Rigorous Integration
Wolfram Alpha suggests using the substitution $\displaystyle u= \frac{x+1}{x- 1}$ precisely what is inside the square root. With that, the integrand becomes just $\displaystyle \sqrt{u}= u^{1/2}$. Now, what about dx?

Well, if $\displaystyle u= \frac{x+1}{x-1}$ then u(x- 1)= ux- u= x+ 1 so ux- x= u+ 1, x(u- 1)= u+1, and $\displaystyle x= \frac{u+1}{u- 1}$. From that, $\displaystyle dx= \frac{(u-1)- (u+1)}{(u-1)^2}du= -\frac{2}{(u-1)^2}du$.

Then $\displaystyle \int\sqrt{\frac{x+1}{x-1}}dx= \int u^{1/2}\left(-\frac{2}{(u-1)^2}\right)du$