$\displaystyle \int \frac{x}{\sqrt{1+2x}}dx$
Or make the sbstitution $\displaystyle u = 1 + 2x \Rightarrow x = \frac{u-1}{2} \, $ and $\displaystyle \, \frac{du}{dx} = 2 \Rightarrow dx = \frac{du}{2}$.
Then the integral becomes
$\displaystyle \int \frac{(u - 1)/2}{\sqrt{u}} \, \frac{du}{2} = \frac{1}{4} \int \frac{u - 1}{\sqrt{u}} \, du$
which can be broken up into simple bits involving powers of u.
mr fantastic, that's the method I used! And I got the right answer, but I made a small arithmetical error in the final simplification ( I added the denominators of 1/4 and 2/3 instead of multiplying them) so the answer didn't check. Sometimes I just see two numbers and instantly do the wrong operation.
Hello, th%$&873!
I prefer to get rid of the radical first . . .
$\displaystyle \int \frac{x}{\sqrt{1+2x}}\,dx$
Let $\displaystyle u \:= \:\sqrt{1+2x}\quad\Rightarrow\quad x \:=\: \frac{1}{2}(u^2-1)\quad\Rightarrow\quad dx \:=\:u\,du$
Substitute: .$\displaystyle \int\frac{\frac{1}{2}(u^2-1)}{u}(u\,du) \;=\;\frac{1}{2}\int(u^2-1)\,du \;=\;\frac{1}{2}\left(\frac{u^3}{3} - u\right) + C \;=\;\frac{1}{6}\,u(u^2-3)+C$
Back-substitute: .$\displaystyle \frac{1}{6}\sqrt{1+2x}\,(1 + 2x - 3) + C \;=\;\frac{1}{6}\sqrt{1+2x}\,(2x-2) + C$
. . Answer: .$\displaystyle \frac{1}{3}\sqrt{1+2x}\,(x-1) + C$