1. ## integration

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

2. Originally Posted by Punch
$\displaystyle \int\frac{x}{(2x+1)^4} dx$
Have you thought of some substitution?

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

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

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

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

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

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

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

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 :

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

6. Originally Posted by yeKciM
solve this than :

$\displaystyle \displaystyle \frac {1}{2} \int \frac {1}{(2x+1)^3} \; dx -\frac {1}{2} \int \frac {1}{(2x+1)^4} \; dx$
i wouldnt mind solving this but i would want to know how u got this

7. partial fractions

8. 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

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

10. i am confused

11. Let u = 2x + 1

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

So the given problem reduces to

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

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

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

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

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

13. Originally Posted by Punch
a little question... $\displaystyle \int\frac{x}{(2x+1)^4} dx$

can i integrate the above by first taking x out and integrate $\displaystyle x\int\frac{1}{(2x+1)^4} dx$?
Yes. You can do it. But you must know the integration by part. Search the text book for the integration by part.

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

15. Originally Posted by Punch
a little question... $\displaystyle \int\frac{x}{(2x+1)^4} dx$

can i integrate the above by first taking x out and integrate $\displaystyle x\int\frac{1}{(2x+1)^4} dx$?
NO! You cannot!

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

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

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

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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