Rigorous Integration

**Cbarker1** · June 5th 2012, 12:09 PM

$\int(\sqrt((x+1)/(x-2))<br />$

**skeeter** · June 5th 2012, 12:20 PM

Originally Posted by Cbarker1

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

integral sqrt((x+1)/(x-2)) dx - Wolfram|Alpha

click on "show steps" ...

**Cbarker1** · June 5th 2012, 12:22 PM

I am confusion how that could be legal in the math world.

**Plato** · June 5th 2012, 12:33 PM

Originally Posted by Cbarker1

I am confusion how that could be legal in the math world.

I am confusion how that could be legal in the math world.

Please tell us what that sentence could possibly mean.

**Cbarker1** · June 5th 2012, 12:42 PM

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?

**HallsofIvy** · June 5th 2012, 05:18 PM

Wolfram Alpha suggests using the substitution $u= \frac{x+1}{x- 1}$ precisely what is inside the square root. With that, the integrand becomes just $\sqrt{u}= u^{1/2}$ . Now, what about dx?

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

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

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