Sorry, the first version was incorrect. This is the correct integral.
$\displaystyle
\int (xe^{2x})/(2x + 1)^2 dx
$
Wolfram tells me that you need to use the Series expansion...
integral[(x*Exp[2*x] )/((2*x + 1)^2)] - Wolfram|Alpha
The thing is, I'm in a chapter in Calculus II that covers several of the integration techniques beyond the basic formulas, such as, integration by parts, trig integrals, trig substitution, partial fractions, L' Hopital's rule, etc. Sequences and Series aren't introduced until the next chapter, which is effectively Calculus III. I suppose this particular integral may have been included accidentally. However, I got close to the answer after integrating by parts several times. Suppose I could have gotten off track somewhere. I may try that technique again and see what I come up with.
Thanks
Try to differentiate $\displaystyle \frac{ e^{2x}}{2x+1} $ and see what would happen !
Consider
$\displaystyle \frac{xe^{2x}}{(2x+1)^2} $
Use these techniques : partial fraction + integration by parts
$\displaystyle = \frac{e^{2x}}{2} \left( \frac{1}{2x+1} - \frac{1}{(2x+1)^2} \right ) $
the integral
$\displaystyle = \frac{1}{2} \int e^{2x} \left( \frac{1}{2x+1} - \frac{1}{(2x+1)^2} \right ) ~dx $
$\displaystyle = \frac{1}{2} \left[ \int e^{2x} \frac{1}{2x+1}~dx - \int e^{2x} \frac{1}{(2x+1)^2}~dx \right ] $
$\displaystyle = \frac{1}{2} \left[ \frac{1}{2} \frac{e^{2x}}{2x+1} + \int e^{2x} \frac{1}{(2x+1)^2}~dx -\int e^{2x} \frac{1}{(2x+1)^2}~dx \right ] $
$\displaystyle = \frac{1}{4} \frac{e^{2x}}{2x+1} $
Thank you. I will study your solution tomorrow. This particular problem is at the end of a section dealing only with integration by parts, i.e., partial fractions come later in chapter. It may be a precursor problem to prepare for the partial fraction material. Either way, I appreciate your solution and, as I mentioned, I will study it.
I appreciate the help!