Integrate e^-3t/(1+e^t)

`int (e^(-3t))/(1+e^t)dt`

Rewrite the integrand:

`(e^(-3t))/(1+e^t)=(e^(-3t)+e^(-2t)-e^(-2t))/(1+e^t)=(e^(-3t)(e^t+1)-e^(-2t))/(e^t+1)`

`=e^(-3t)-(e^(-2t))/(e^t+1)=e^(-3t)+(-e^(-2t)+e^(-t)-e^(-t))/(e^t+1)=e^(-3t)+(-e^(-2t)(e^t+1)+e^(-t))/(e^t+1)`

`=e^(-3t)-e^(-2t)+(e^(-t))/(e^t+1)`

Looking just at the last term:

`(e^(-t))/(e^t+1)=(e^(-t)+1-1)/(e^t+1)=(e^(-t)(e^t+1)-1)/(e^t+1)=e^(-t)-1/(e^t+1)`

So:

`int (e^(-3t))/(1+e^t)dt=int (e^(-3t)-e^(-2t)+e^(-t)-1/(e^t+1))dt`

The last term in this integral can be rewritten:

`1/(e^t+1)*e^(-t)/e^(-t)=(e^(-t))/(1+e^(-t))`

Integrating term by term we get:

`int (e^(-3t))/(1+e^t)dt=-1/3e^(-3t)+1/2e^(-2t)-e^(-t)+ln(e^(-t)+1)+C`

Comments

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