How can we compute $\int \frac{te^t}{(1+t)^2}dt$?

Question

This integral seems to be simple to calculate, but i cant. How can we compute $\int \frac{te^t}{(1+t)^2}dt$?

Answer 1

Use integration by parts, to get:

$$\mathcal{I}\left(t\right)=\int\frac{te^t}{\left(1+t\right)^2}\space\text{d}t=-\frac{te^t}{1+t}+\int e^t\space\text{d}t$$

Integration by parts:

$$\int\text{f}\left(t\right)\cdot\text{g}'\left(t\right)\space\text{d}t=\text{f}\left(t\right)\cdot\text{g}\left(t\right)-\int\text{f}'\left(t\right)\cdot\text{g}\left(t\right)\space\text{d}t$$

Answer 2

In addition to Jan's simple answer, maybe you can do this as well:

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Let $u=(t+1) $ then our integral becomes $$I=\frac {1}{e } \int \frac {(u-1)e^u }{u^2} =\frac {1}{e}[\int \frac{e^u}{u} du -\int \frac {e^u}{u^2} du] $$ Integrating $\frac {e^u}{u^2} $ By parts where $f=e^u $ and $g'=\frac {1}{u^2} $, we get, $$\int \frac {e^u}{u^2} =\int \frac {e^u}{u} - \frac {e^u}{u} $$ Rearranging, we get the answer. Hope it helps.