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Math Help - Can I simplify the following function

  1. #1
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    Can I simplify the following function

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

    so that the antiderivative will be x+e^t+C
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    Quote Originally Posted by cammywhite View Post
    Can I simplify \int \frac {e^t}{e^t+1}dt into \int 1+e^t dt

    so that the antiderivative will be x+e^t+C
    um, no. this is a substitution problem. let u = 1 + e^t
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    Quote Originally Posted by cammywhite View Post
    Can I simplify \int \frac {e^t}{e^t+1}dt into \int 1+e^t dt

    so that the antiderivative will be x+e^t+C
    No.

    Use substitution. Let  u = e^t + 1

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

     du = e^t dt

    Therefore integral becomes:

     \int \frac{du}{u}
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  4. #4
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    Quote Originally Posted by Mush View Post
    No.

    Use substitution. Let  u = e^t + 1

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

     du = e^t dt

    Therefore integral becomes:

     \int \frac{du}{u}
    so it will be \frac {e^t+x}{e^t+1}
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    is up to his old tricks again! Jhevon's Avatar
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    *sigh* please review your integration rules: \int \frac 1x~dx = \ln |x| + C
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    Quote Originally Posted by Jhevon View Post
    *sigh* please review your integration rules: \int \frac 1x~dx = \ln |x| + C
    I just have problem with the substituting part...

    Let u=e^t+1

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

    am i correct???
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    Quote Originally Posted by cammywhite View Post
    I just have problem with the substituting part...

    Let u=e^t+1

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

    am i correct???
    what are you doing?! are you reading the posts made here?

    picking up where Mush left off: we have

    \int \frac 1u~du

    applying the rule i just gave you, we get

    \ln |u| + C

    now back-substitute for u, we get:

    \ln |e^t + 1| + C

    please re-read posts #3 and #5 carefully
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  8. #8
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    Maybe some colour would help.

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

    So your integral: \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.
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    what are you doing?! are you reading the posts made here?

    picking up where Mush left off: we have

    \int \frac 1u~du

    applying the rule i just gave you, we get

    \ln |u| + C

    now back-substitute for u, we get:

    \ln |e^t + 1| + C

    please re-read posts #3 and #5 carefully

    So in otherwords the derivative of \ln |e^t + 1| + C is \frac {e^t}{e^t+1}?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cammywhite View Post
    So in otherwords the derivative of \ln |e^t + 1| + C is \frac {e^t}{e^t+1}?
    yes. by the chain rule: \frac d{dx} \ln u = \frac {u'}u

    here u is a function of x
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