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Math Help - Can't work out why u=e^x means u^2 becomes e^x ?

  1. #1
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    Can't work out why u=e^x means u^2 becomes e^x ?

    Question says Show that the substitution u = e^x converts

    \int \frac{2 + ln u}{u^2} du into

    \int \frac{2 + x}{e^e} dx

    I understand how ln e^x = xlne = x .

    But Why does u^2 become e^x ? Would it become e^x squared?

    Angus
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Can't work out why u=e^x means u^2 becomes e^x ?

    Hi angypangy!


    Yes, u^2 becomes (e^x)^2.

    But du becomes e^x dx.

    The e^x cancels against the denominator.
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  3. #3
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    Re: Can't work out why u=e^x means u^2 becomes e^x ?

    Quote Originally Posted by angypangy View Post
    Question says Show that the substitution u = e^x converts

    \int \frac{2 + ln u}{u^2} du into

    \int \frac{2 + x}{e^e} dx

    I understand how ln e^x = xlne = x .

    But Why does u^2 become e^x ? Would it become e^x squared?

    Angus

    From the laws of logarithms \ln(e^x)=x \ln(e), and by definition \ln(e)=1

    CB
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    Re: Can't work out why u=e^x means u^2 becomes e^x ?

    Quote Originally Posted by ILikeSerena View Post
    Hi angypangy!


    Yes, u^2 becomes (e^x)^2.

    But du becomes e^x dx.

    The e^x cancels against the denominator.

    I am not getting this. Could you please work through how to do it.

    This is where I get to.

    \int \frac{2 + x}{e^e} dx

    Then let u = e^x

    \frac{du}{dx} (u = e^x) = e^x

    ie du = e^x dx

    \int \frac{(2 + x)(u)}{u} du

    and

    Then what?
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  5. #5
    Super Member ILikeSerena's Avatar
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    Re: Can't work out why u=e^x means u^2 becomes e^x ?

    Quote Originally Posted by angypangy View Post
    I am not getting this. Could you please work through how to do it.

    This is where I get to.

    \int \frac{2 + x}{e^e} dx

    Then let u = e^x

    \frac{du}{dx} (u = e^x) = e^x

    ie du = e^x dx

    \int \frac{(2 + x)(u)}{u} du

    and

    Then what?
    Starting from: du = e^x dx

    This means dx = {du \over e^x} = {du \over u}

    So:

    \int \frac{(2 + x)}{u} {du \over u} = \int \frac{(2 + \ln u)}{u^2} du
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