1. ## Please give me some hints:

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

2. 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

3. ## Substitution

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.

4. Deleted

5. ## thanks

thanks a lot guys!

6. 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