$\displaystyle \int\frac{x}{(2x+1)^4} dx$
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
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.
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.
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.
$\displaystyle \displaystyle \frac{x}{(2x + 1)^4} = \frac{1}{2} \cdot \frac{2x}{(2x + 1)^4} = \frac{1}{2} \cdot \frac{(2x + 1) - 1}{(2x + 1)^4}$
$\displaystyle \displaystyle = \frac{1}{2} \left( \frac{2x + 1}{(2x + 1)^4} - \frac{1}{(2x + 1)^4} \right) = \frac{1}{2} \left( \frac{1}{(2x + 1)^3} - \frac{1}{(2x + 1)^4} \right)$
This is misleading at best and wrong at worst.