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

$$\int \frac{t}{e^{-2t}}dt$$

Answered by TheSupreme last updated on 08/Feb/22

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

Commented by alephzero last updated on 08/Feb/22

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

Answered by alephzero last updated on 08/Feb/22

$$\int \frac{t}{e^{-2t}}dt=\int \frac{t}{\frac{1}{e^{2t}}}dt=\int {te}^{2t}dt=$$$$=\int td\left(\frac{e^{2t}}{2}\right)=\frac{{te}^{2t}}{2}-\int \frac{e^{2t}}{2}dt=$$$$=\frac{{te}^{2t}}{2}-\frac{1}{2}\int e^{2t}dt=\frac{{te}^{2t}}{2}-\frac{1}{2}\left(\frac{e^{2t}}{2}\right)=$$$$=\frac{{te}^{2t}}{2}-\frac{e^{2t}}{4}=\frac{2{te}^{2t}-e^{2t}}{4}=$$$$=\frac{e^{2t}(2t-1)}{4}$$$$\Rightarrow \int \frac{t}{e^{-2t}}dt=\frac{e^{2t}(2t-1)}{4}+C$$

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