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

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

$$ \\ $$$$\int\frac{{x}\:{e}^{\mathrm{2}{x}} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\:\:\:{please}\:{help} \\ $$

Answered by thfchristopher last updated on 07/Jun/22

∫((x e^(2x) )/((2x+1)^2 ))dx =−(1/2)∫xe^(2x) d(2x+1)^(-1) =−(1/2)[((xe^(2x) )/(2x+1))−∫((d(xe^(2x) ))/(2x+1))] =−(1/2)[((xe^(2x) )/(2x+1))−∫((e^(2x) +2xe^(2x) )/(2x+1))dx] =−(1/2)[((xe^(2x) )/(2x+1))−∫e^(2x) dx] =−(1/2)[((xe^(2x) )/(2x+1))−(e^(2x) /2)]+C =(e^(2x) /(4(2x+1)))+C

$$\int\frac{{x}\:{e}^{\mathrm{2}{x}} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int{xe}^{\mathrm{2}{x}} {d}\left(\mathrm{2}{x}+\mathrm{1}\right)^{-\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{xe}^{\mathrm{2}{x}} }{\mathrm{2}{x}+\mathrm{1}}−\int\frac{{d}\left({xe}^{\mathrm{2}{x}} \right)}{\mathrm{2}{x}+\mathrm{1}}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{xe}^{\mathrm{2}{x}} }{\mathrm{2}{x}+\mathrm{1}}−\int\frac{{e}^{\mathrm{2}{x}} +\mathrm{2}{xe}^{\mathrm{2}{x}} }{\mathrm{2}{x}+\mathrm{1}}{dx}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{xe}^{\mathrm{2}{x}} }{\mathrm{2}{x}+\mathrm{1}}−\int{e}^{\mathrm{2}{x}} {dx}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{xe}^{\mathrm{2}{x}} }{\mathrm{2}{x}+\mathrm{1}}−\frac{{e}^{\mathrm{2}{x}} }{\mathrm{2}}\right]+{C} \\ $$$$=\frac{{e}^{\mathrm{2}{x}} }{\mathrm{4}\left(\mathrm{2}{x}+\mathrm{1}\right)}+{C} \\ $$

Commented by Kalebwizeman last updated on 08/Jun/22

thank you very much Chris

$${thank}\:{you}\:{very}\:{much}\:{Chris} \\ $$

Commented by Kalebwizeman last updated on 08/Jun/22

please may I ask sir what informed your choice of u in the IBP? is it LIATE? is xe^(2x) algebraic? please

$${please}\:{may}\:{I}\:{ask}\:{sir}\:{what}\:{informed}\:{your}\:{choice}\:{of}\:{u}\:{in}\:{the}\:{IBP}? \\ $$$${is}\:{it}\:{LIATE}?\:{is}\:{xe}^{\mathrm{2}{x}} \:{algebraic}?\:{please} \\ $$

Commented by thfchristopher last updated on 08/Jun/22

I don′t know the LIATE rule but just simply want to make ∫udv easily and directly. If you put xe^(2x) into dv, you should know what the thing was differentiated to become xe^(2x) . (2x+1)^(−2) can be thought quickly as it come from −(1/2)(2x+1)^(−1) . Therefore I can make −(1/2)∫xe^(2x) d(2x+1)^(−1) and continue my evaluation. Certainly if I stuck in the later progress, I may try putting xe^(2x) , e^(2x) or x and look whether the answer can be finally derived. Hope it can help.

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{LIATE}\:\mathrm{rule}\:\mathrm{but}\:\mathrm{just}\:\mathrm{simply} \\ $$$$\mathrm{want}\:\mathrm{to}\:\mathrm{make}\:\int{udv}\:\mathrm{easily}\:\mathrm{and}\:\mathrm{directly}. \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{put}\:{xe}^{\mathrm{2}{x}} \:\mathrm{into}\:{dv},\:\mathrm{you}\:\mathrm{should}\:\mathrm{know}\:\mathrm{what} \\ $$$$\mathrm{the}\:\mathrm{thing}\:\mathrm{was}\:\mathrm{differentiated}\:\mathrm{to}\:\mathrm{become}\:{xe}^{\mathrm{2}{x}} . \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right)^{−\mathrm{2}} \:\mathrm{can}\:\mathrm{be}\:\mathrm{thought}\:\mathrm{quickly}\:\mathrm{as}\:\mathrm{it}\:\mathrm{come} \\ $$$$\mathrm{from}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{−\mathrm{1}} .\:\mathrm{Therefore}\:\mathrm{I}\:\mathrm{can}\:\mathrm{make} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int{xe}^{\mathrm{2}{x}} {d}\left(\mathrm{2}{x}+\mathrm{1}\right)^{−\mathrm{1}} \:\mathrm{and}\:\mathrm{continue}\:\mathrm{my} \\ $$$$\mathrm{evaluation}.\:\mathrm{Certainly}\:\mathrm{if}\:\mathrm{I}\:\mathrm{stuck}\:\mathrm{in}\:\mathrm{the}\:\mathrm{later} \\ $$$$\mathrm{progress},\:\mathrm{I}\:\mathrm{may}\:\mathrm{try}\:\mathrm{putting}\:{xe}^{\mathrm{2}{x}} ,\:{e}^{\mathrm{2}{x}} \:\mathrm{or}\:{x} \\ $$$$\mathrm{and}\:\mathrm{look}\:\mathrm{whether}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{can}\:\mathrm{be}\:\mathrm{finally} \\ $$$$\mathrm{derived}. \\ $$$$\mathrm{Hope}\:\mathrm{it}\:\mathrm{can}\:\mathrm{help}. \\ $$

Commented by Kalebwizeman last updated on 08/Jun/22

really helpful. Thank you immensely. More knowledge to you

$${really}\:{helpful}.\:{Thank}\:{you}\:{immensely}.\:{More}\:{knowledge}\:{to}\:{you} \\ $$$$ \\ $$

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