integration

**Punch** · August 28th 2010, 06:20 AM

$\int\frac{x}{(2x+1)^4} dx$

**Isomorphism** · August 28th 2010, 06:21 AM

Originally Posted by Punch

$\int\frac{x}{(2x+1)^4} dx$

Have you thought of some substitution?

**yeKciM** · August 28th 2010, 06:25 AM

Originally Posted by Punch

$\int\frac{x}{(2x+1)^4} dx$

$\displaystyle \int \frac {x}{(2x+1)^4} \; dx =\frac {1}{2} \int \frac {1}{(2x+1)^3} \; dx -\frac {1}{2} \int \frac {1}{(2x+1)^4} \; dx$

use substitution $u = 2x+1$ so $du=2dx$ and so on

**Punch** · August 28th 2010, 06:26 AM

sorry, but i dont integrate with substituting. I learnt another method that is... let me give an example

$\int\frac{1}{(2x+1)^4} dx=-\frac{1}{6(2x+1)^3}$

i add the power by 1 and divide the whole function by the new power, and divide it once again by the derivative.

however, this function has x in both the denominator and the numerator, which is what makes it tough for me

**yeKciM** · August 28th 2010, 06:32 AM

Originally Posted by Punch

sorry, but i dont integrate with substituting. I learnt another method that is... let me give an example

$\int\frac{1}{(2x+1)^4} dx=-\frac{1}{6(2x+1)^3}$

i add the power by 1 and divide the whole function by the new power, and divide it once again by the derivative.

however, this function has x in both the denominator and the numerator, which is what makes it tough for me

solve this than :

$\displaystyle \frac {1}{2} \int \frac {1}{(2x+1)^3} \; dx -\frac {1}{2} \int \frac {1}{(2x+1)^4} \; dx$

**Punch** · August 28th 2010, 06:33 AM

Originally Posted by yeKciM

solve this than :

$\displaystyle \frac {1}{2} \int \frac {1}{(2x+1)^3} \; dx -\frac {1}{2} \int \frac {1}{(2x+1)^4} \; dx$

i wouldnt mind solving this but i would want to know how u got this

**yeKciM** · August 28th 2010, 06:36 AM

partial fractions

**Punch** · August 28th 2010, 06:55 AM

it isnt possible that i would need to do partial fractions as i have not learnt how to turn such an equation into partial fractions

**HallsofIvy** · August 28th 2010, 09:43 AM

No, there are no "partial fractions" involved in these problem. In $\int\frac{x}{(2x+1)^4} \, dx$ , $\int \frac{1}{2x+1}^3 \, dx$ , and $\int \frac{1}{(2x+1)^4} \, dx$ , let u= 2x+ 1.

**Punch** · August 28th 2010, 06:15 PM

i am confused

**sa-ri-ga-ma** · August 28th 2010, 09:38 PM

Let u = 2x + 1

Then $x = \frac{u - 1}{2}$ and dx = du/2.

So the given problem reduces to

$\frac{1}{2}(\frac{u-1}{u^4})(\frac{du}{2})$

$\frac{1}{2}(\frac{u}{u^4}-\frac{1}{u^4})(\frac{du}{2})$

Now find the integration by your method and then substitute the value of u to get the result.

**Punch** · August 28th 2010, 09:57 PM

a little question... $\int\frac{x}{(2x+1)^4} dx$

can i integrate the above by first taking x out and integrate $x\int\frac{1}{(2x+1)^4} dx$ ?

**sa-ri-ga-ma** · August 28th 2010, 10:15 PM

Originally Posted by Punch

a little question... $\int\frac{x}{(2x+1)^4} dx$

can i integrate the above by first taking x out and integrate $x\int\frac{1}{(2x+1)^4} dx$ ?

Yes. You can do it. But you must know the integration by part. Search the text book for the integration by part.

**Punch** · August 28th 2010, 10:17 PM

Okay, i have no problem integrating $x\int\frac{1}{(2x+1)^4} dx$ ... but my doubt is that in such a way, $x$ remains unintegrated!!!

**mr fantastic** · August 28th 2010, 10:32 PM

Originally Posted by Punch

a little question... $\int\frac{x}{(2x+1)^4} dx$

can i integrate the above by first taking x out and integrate $x\int\frac{1}{(2x+1)^4} dx$ ?

NO! You cannot!

You have been shown how to do this question: The integrand can be written as $\displaystyle \frac{1}{2} \left( \frac{1}{(2x + 1)^3} - \frac{1}{(2x + 1)^4} \right)$ (see below) and you claim in an earlier post to be able to integrate expressions like these. So there shouldn't be any problem.

$\displaystyle \frac{x}{(2x + 1)^4} = \frac{1}{2} \cdot \frac{2x}{(2x + 1)^4} = \frac{1}{2} \cdot \frac{(2x + 1) - 1}{(2x + 1)^4}$

$\displaystyle = \frac{1}{2} \left( \frac{2x + 1}{(2x + 1)^4} - \frac{1}{(2x + 1)^4} \right) = \frac{1}{2} \left( \frac{1}{(2x + 1)^3} - \frac{1}{(2x + 1)^4} \right)$

Originally Posted by sa-ri-ga-ma

Yes. You can do it. But you must know the integration by part. Search the text book for the integration by part.

This is misleading at best and wrong at worst.

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