1. ## \int \frac{x}{\sqrt{1+2x}}dx

$\int \frac{x}{\sqrt{1+2x}}dx$

2. integration by parts:

$\int {\frac{x}
{{\sqrt {1 + 2x} }}dx} = x\sqrt {1 + 2x} - \int {\sqrt {1 + 2x} dx = } x\sqrt {1 + 2x} - \frac{1}
{3}\left( {1 + 2x} \right)^{\frac{3}
{2}}$

3. Originally Posted by Peritus
integration by parts:

$\int {\frac{x}
{{\sqrt {1 + 2x} }}dx} = x\sqrt {1 + 2x} - \int {\sqrt {1 + 2x} dx = } x\sqrt {1 + 2x} - \frac{1}
{3}\left( {1 + 2x} \right)^{\frac{3}
{2}}$
Or make the sbstitution $u = 1 + 2x \Rightarrow x = \frac{u-1}{2} \,$ and $\, \frac{du}{dx} = 2 \Rightarrow dx = \frac{du}{2}$.

Then the integral becomes

$\int \frac{(u - 1)/2}{\sqrt{u}} \, \frac{du}{2} = \frac{1}{4} \int \frac{u - 1}{\sqrt{u}} \, du$

which can be broken up into simple bits involving powers of u.

4. 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.

5. Originally Posted by th%$&873 mr fantastic, that's the method I used! And I got the right answer, but I made a small arithmetical error in the final simplification ( I added the denominators of 1/4 and 2/3 instead of multiplying them) so the answer didn't check. Sometimes I just see two numbers and instantly do the wrong operation. Became a pro at the abuse of substitution and you will save yourself a lot of headaches! 6. Hello, th%$&873!

I prefer to get rid of the radical first . . .

$\int \frac{x}{\sqrt{1+2x}}\,dx$

Let $u \:= \:\sqrt{1+2x}\quad\Rightarrow\quad x \:=\: \frac{1}{2}(u^2-1)\quad\Rightarrow\quad dx \:=\:u\,du$

Substitute: . $\int\frac{\frac{1}{2}(u^2-1)}{u}(u\,du) \;=\;\frac{1}{2}\int(u^2-1)\,du \;=\;\frac{1}{2}\left(\frac{u^3}{3} - u\right) + C \;=\;\frac{1}{6}\,u(u^2-3)+C$

Back-substitute: . $\frac{1}{6}\sqrt{1+2x}\,(1 + 2x - 3) + C \;=\;\frac{1}{6}\sqrt{1+2x}\,(2x-2) + C$

. . Answer: . $\frac{1}{3}\sqrt{1+2x}\,(x-1) + C$

7. Originally Posted by th%\$&873
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.
Welcome to the club, chum!