Originally Posted by
TwistedOne151 topsquark, if we let $\displaystyle y = b + \sqrt{x}$, then we have
$\displaystyle \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$
$\displaystyle dx = 2\sqrt{x} dy = 2(y - b) dy$, right?
(We could also write $\displaystyle y = b + \sqrt{x} \implies x = (y - b)^2 $, so $\displaystyle \dfrac{dx}{dy} = 2(y - b)$, and again, $\displaystyle dx = 2(y - b) dy$)
Thus we have
$\displaystyle \int \frac{2\sqrt{x}}{b + \sqrt{x}}~dx = \int \frac{2(y - b)}{b + (y - b)} \cdot 2(y - b)~dy$
$\displaystyle = 4 \int \frac{(y - b)^2}{y}~dy = 4 \int (y - 2b + \frac{b^2}{y})~dy$
For a noticeably different result.
--Kevin C.