Please give me some hints:

**Tuugii** · December 3rd 2007, 10:39 PM

I need to find the integral of:

(2*x^(1/2))/(b+x^(1/2)), where b is a constant real number.

thanks a lot,
T

**Jhevon** · December 3rd 2007, 10:51 PM

Originally Posted by Tuugii

I need to find the integral of:

(2*x^(1/2))/(b+x^(1/2)), where b is a constant real number.

thanks a lot,
T

there may be an easier way i'm not seeing, but it's 3am, cut me some slack. factor out the $x^{1/2}$ from the denominator, then do a substitution of $u = 2x^{1/2}$ . simplify as much as possible, then do another substitution of $t = 2b + u$ . the resulting integral should be nice

**TwistedOne151** · December 4th 2007, 07:38 AM

topsquark, if we let $y = b + \sqrt{x}$ , then we have
$\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$
$dx = 2\sqrt{x} dy = 2(y - b) dy$ , right?

(We could also write $y = b + \sqrt{x} \implies x = (y - b)^2$ , so $\dfrac{dx}{dy} = 2(y - b)$ , and again, $dx = 2(y - b) dy$ )

Thus we have
$\int \frac{2\sqrt{x}}{b + \sqrt{x}}~dx = \int \frac{2(y - b)}{b + (y - b)} \cdot 2(y - b)~dy$
$= 4 \int \frac{(y - b)^2}{y}~dy = 4 \int (y - 2b + \frac{b^2}{y})~dy$

For a noticeably different result.

--Kevin C.

**colby2152** · December 4th 2007, 07:41 AM

Deleted

**Tuugii** · December 4th 2007, 07:58 AM

thanks a lot guys!

**topsquark** · December 4th 2007, 10:19 AM

Originally Posted by TwistedOne151

topsquark, if we let $y = b + \sqrt{x}$ , then we have
$\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$
$dx = 2\sqrt{x} dy = 2(y - b) dy$ , right?

(We could also write $y = b + \sqrt{x} \implies x = (y - b)^2$ , so $\dfrac{dx}{dy} = 2(y - b)$ , and again, $dx = 2(y - b) dy$ )

Thus we have
$\int \frac{2\sqrt{x}}{b + \sqrt{x}}~dx = \int \frac{2(y - b)}{b + (y - b)} \cdot 2(y - b)~dy$
$= 4 \int \frac{(y - b)^2}{y}~dy = 4 \int (y - 2b + \frac{b^2}{y})~dy$

For a noticeably different result.

--Kevin C.

Yup. That's what I get for writing things down in the margin of my paper! Thanks for catching that.

-Dan

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