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

**th%$&873** · February 20th 2008, 03:50 PM

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

**Peritus** · February 20th 2008, 04:41 PM

integration by parts:

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

**mr fantastic** · February 20th 2008, 05:01 PM

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.

**th%$&873** · February 21st 2008, 11:19 AM

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.

**colby2152** · February 21st 2008, 12:11 PM

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!

**Soroban** · February 21st 2008, 12:30 PM

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

**mr fantastic** · February 21st 2008, 02:19 PM

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!

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