∫((x e^(2x) )/((2x+1)^2 ))dx please help


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Question Number 171096 by Kalebwizeman last updated on 07/Jun/22

$\int \frac{x e^{2 x}}{{(2 x + 1)}^{2}} d x p l e a s e h e l p$
	Answered by thfchristopher last updated on 07/Jun/22

	$\int \frac{x e^{2 x}}{{(2 x + 1)}^{2}} d x$ $= - \frac{1}{2} \int {x e}^{2 x} d {(2 x + 1)}^{- 1}$ $= - \frac{1}{2} [\frac{{x e}^{2 x}}{2 x + 1} - \int \frac{d ({x e}^{2 x})}{2 x + 1}]$ $= - \frac{1}{2} [\frac{{x e}^{2 x}}{2 x + 1} - \int \frac{e^{2 x} + 2 {x e}^{2 x}}{2 x + 1} d x]$ $= - \frac{1}{2} [\frac{{x e}^{2 x}}{2 x + 1} - \int e^{2 x} d x]$ $= - \frac{1}{2} [\frac{{x e}^{2 x}}{2 x + 1} - \frac{e^{2 x}}{2}] + C$ $= \frac{e^{2 x}}{4 (2 x + 1)} + C$
	Commented by Kalebwizeman last updated on 08/Jun/22

	$t h a n k y o u v e r y m u c h C h r i s$
	Commented by Kalebwizeman last updated on 08/Jun/22

	$p l e a s e m a y I a s k s i r w h a t i n f o r m e d y o u r c h o i c e o f u i n t h e I B P ?$ $i s i t L I A T E ? i s {x e}^{2 x} a l g e b r a i c ? p l e a s e$
	Commented by thfchristopher last updated on 08/Jun/22

	$I {don}^{'} t know the LIATE rule but just simply$ $want to make \int u d v easily and directly .$ $If you put {x e}^{2 x} into d v, you should know what$ $the thing was differentiated to become {x e}^{2 x} .$ ${(2 x + 1)}^{- 2} can be thought quickly as it come$ $from - \frac{1}{2} {(2 x + 1)}^{- 1} . Therefore I can make$ $- \frac{1}{2} \int {x e}^{2 x} d {(2 x + 1)}^{- 1} and continue my$ $evaluation . Certainly if I stuck in the later$ $progress, I may try putting {x e}^{2 x}, e^{2 x} or x$ $and look whether the answer can be finally$ $derived .$ $Hope it can help .$
	Commented by Kalebwizeman last updated on 08/Jun/22

	$r e a l l y h e l p f u l . T h a n k y o u i m m e n s e l y . M o r e k n o w l e d g e t o y o u$
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