# Can I simplify the following function

• Apr 5th 2009, 05:59 PM
cammywhite
Can 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 PM
Jhevon
Quote:

Originally Posted by cammywhite
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 $\displaystyle x+e^t+C$

um, no. this is a substitution problem. let $\displaystyle u = 1 + e^t$
• Apr 5th 2009, 06:03 PM
Mush
Quote:

Originally Posted by cammywhite
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 $\displaystyle x+e^t+C$

No.

Use substitution. Let $\displaystyle u = e^t + 1$

$\displaystyle \frac{du}{dt} = e^t$

$\displaystyle du = e^t dt$

Therefore integral becomes:

$\displaystyle \int \frac{du}{u}$
• Apr 5th 2009, 06:09 PM
cammywhite
Quote:

Originally Posted by Mush
No.

Use substitution. Let $\displaystyle u = e^t + 1$

$\displaystyle \frac{du}{dt} = e^t$

$\displaystyle du = e^t dt$

Therefore integral becomes:

$\displaystyle \int \frac{du}{u}$

so it will be $\displaystyle \frac {e^t+x}{e^t+1}$
• Apr 5th 2009, 06:26 PM
Jhevon
*sigh* please review your integration rules: $\displaystyle \int \frac 1x~dx = \ln |x| + C$
• Apr 5th 2009, 06:37 PM
cammywhite
Quote:

Originally Posted by Jhevon
*sigh* please review your integration rules: $\displaystyle \int \frac 1x~dx = \ln |x| + C$

I just have problem with the substituting part...

Let $\displaystyle u=e^t+1$

so$\displaystyle \int \frac {e^t}{e^t+1}dt$ will equal to $\displaystyle \int \frac {u-1}{u}dt$

am i correct???
• Apr 5th 2009, 06:40 PM
Jhevon
Quote:

Originally Posted by cammywhite
I just have problem with the substituting part...

Let $\displaystyle u=e^t+1$

so$\displaystyle \int \frac {e^t}{e^t+1}dt$ will equal to $\displaystyle \int \frac {u-1}{u}dt$

am i correct???

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:

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

• Apr 5th 2009, 06:42 PM
o_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 PM
cammywhite
Quote:

Originally Posted by Jhevon

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:

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

So in otherwords the derivative of $\displaystyle \ln |e^t + 1| + C$ is $\displaystyle \frac {e^t}{e^t+1}?$
So in otherwords the derivative of $\displaystyle \ln |e^t + 1| + C$ is $\displaystyle \frac {e^t}{e^t+1}?$
yes. by the chain rule: $\displaystyle \frac d{dx} \ln u = \frac {u'}u$