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Question Number 165794 by SANOGO last updated on 08/Feb/22

∫(t/e^(−2t) )dt

$$\int\frac{{t}}{{e}^{−\mathrm{2}{t}} }{dt} \\ $$

Answered by TheSupreme last updated on 08/Feb/22

∫te^(2t) =((te^t )/2)−(1/4)e^(2t) +c

$$\int{te}^{\mathrm{2}{t}} =\frac{{te}^{{t}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}{e}^{\mathrm{2}{t}} +{c} \\ $$$$ \\ $$

Commented by alephzero last updated on 08/Feb/22

Reonsider ((te^t )/2). Are you sure there  is only t?

$$\mathrm{Reonsider}\:\frac{{te}^{\boldsymbol{{t}}} }{\mathrm{2}}.\:\mathrm{Are}\:\mathrm{you}\:\mathrm{sure}\:\mathrm{there} \\ $$$$\mathrm{is}\:\mathrm{only}\:\boldsymbol{{t}}? \\ $$

Answered by alephzero last updated on 08/Feb/22

∫(t/e^(−2t) )dt = ∫((   t  )/(1/e^(2t) ))dt = ∫te^(2t) dt =  = ∫t d((e^(2t) /2)) = ((te^(2t) )/2)−∫(e^(2t) /2)dt =  = ((te^(2t) )/2)−(1/2)∫e^(2t) dt = ((te^(2t) )/2)−(1/2)((e^(2t) /2)) =  = ((te^(2t) )/2)−(e^(2t) /4) = ((2te^(2t) −e^(2t) )/4) =  = ((e^(2t) (2t−1))/4)   ⇒ ∫(t/e^(−2t) )dt = ((e^(2t) (2t−1))/4) + C

$$\int\frac{{t}}{{e}^{−\mathrm{2}{t}} }{dt}\:=\:\int\frac{\:\:\:{t}\:\:}{\frac{\mathrm{1}}{{e}^{\mathrm{2}{t}} }}{dt}\:=\:\int{te}^{\mathrm{2}{t}} {dt}\:= \\ $$$$=\:\int{t}\:{d}\left(\frac{{e}^{\mathrm{2}{t}} }{\mathrm{2}}\right)\:=\:\frac{{te}^{\mathrm{2}{t}} }{\mathrm{2}}−\int\frac{{e}^{\mathrm{2}{t}} }{\mathrm{2}}{dt}\:= \\ $$$$=\:\frac{{te}^{\mathrm{2}{t}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{\mathrm{2}{t}} {dt}\:=\:\frac{{te}^{\mathrm{2}{t}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{e}^{\mathrm{2}{t}} }{\mathrm{2}}\right)\:= \\ $$$$=\:\frac{{te}^{\mathrm{2}{t}} }{\mathrm{2}}−\frac{{e}^{\mathrm{2}{t}} }{\mathrm{4}}\:=\:\frac{\mathrm{2}{te}^{\mathrm{2}{t}} −{e}^{\mathrm{2}{t}} }{\mathrm{4}}\:= \\ $$$$=\:\frac{{e}^{\mathrm{2}{t}} \left(\mathrm{2}{t}−\mathrm{1}\right)}{\mathrm{4}}\: \\ $$$$\Rightarrow\:\int\frac{{t}}{{e}^{−\mathrm{2}{t}} }{dt}\:=\:\frac{{e}^{\mathrm{2}{t}} \left(\mathrm{2}{t}−\mathrm{1}\right)}{\mathrm{4}}\:+\:{C} \\ $$

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