Can I simplify$\displaystyle \int \frac {e^t}{e^t+1}dt$ into $\displaystyle \int 1+e^t dt$

so that the antiderivative will be $\displaystyle x+e^t+C$

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- Apr 5th 2009, 05:59 PMcammywhiteCan I simplify the following function
Can I simplify$\displaystyle \int \frac {e^t}{e^t+1}dt$ into $\displaystyle \int 1+e^t dt$

so that the antiderivative will be $\displaystyle x+e^t+C$ - Apr 5th 2009, 06:02 PMJhevon
- Apr 5th 2009, 06:03 PMMush
- Apr 5th 2009, 06:09 PMcammywhite
- Apr 5th 2009, 06:26 PMJhevon
*sigh* please review your integration rules: $\displaystyle \int \frac 1x~dx = \ln |x| + C$

- Apr 5th 2009, 06:37 PMcammywhite
- Apr 5th 2009, 06:40 PMJhevon
what are you doing?! are you reading the posts made here?

picking up where**Mush**left off: we have

$\displaystyle \int \frac 1u~du$

applying the rule i just gave you, we get

$\displaystyle \ln |u| + C$

now back-substitute for $\displaystyle u$, we get:

$\displaystyle \ln |e^t + 1| + C$

please re-read posts #3 and #5 carefully - Apr 5th 2009, 06:42 PMo_O
Maybe some colour would help.

$\displaystyle {\color{red}u = e^t + 1} \ \Rightarrow \ {\color{blue}du = e^t dt}$

So your integral: $\displaystyle \int \frac{{\color{blue}e^t dt}}{{\color{red}e^t + 1}} = \int \frac{{\color{blue}du}}{{\color{red}u}}$

which is a standard one. - Apr 5th 2009, 06:48 PMcammywhite
- Apr 5th 2009, 06:58 PMJhevon