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Math Help - U substitution question

  1. #1
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    U substitution question

    Let u = e6x + 9. Then du = 6e6x dx and e6x dx =
    1
    6
    du, so

    e6x
    e6x + 9
    dx
    =
    1
    u
    1
    6
    du
    =
    1
    6
    ln|u| + C =
    1
    6
    ln(e6x + 9) + C.



    Is this correct?


    edit: my syntax is messed up sorry I will fix
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  2. #2
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    Re: U substitution question

    ∫▒〖e^6x/(e^6x+9) dx〗 let e^(6x )=u Therefore 6e^6x dx=du
    hence
    ∫▒〖e^6x/(e^6x+9) dx〗 = 1/6 ∫▒〖1/(u+9) du〗 Now you can finish it
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  3. #3
    Senior Member x3bnm's Avatar
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    Re: U substitution question

    \int \frac{e^{6x}}{e^{6x}+9}\,\, dx

    Let u = e^{6x} + 9\,\,\therefore du = 6e^{6x}\,\,dx

    Now plugging in u and du we get:

    \begin{align*}\int \frac{e^{6x}}{e^{6x}+9}\,\, dx=& \int \frac{1}{6u}\,\,du \\=& \frac{1}{6} \ln|u| + C\\=& \frac{1}{6}\ln| e^{6x}+9| + C.....\text{[by plugging in value of } u\text{]}\end{align*}

    \int \frac{e^{6x}}{e^{6x}+9}\,\, dx = \frac{1}{6}\ln| e^{6x}+9| + C.....\text{[Answer]}

    You can check the answer here:

    int (e^(6x))/(e^(6x)+9) - Wolfram|Alpha
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  4. #4
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    Re: U substitution question

    Quote Originally Posted by x3bnm View Post
    \int \frac{e^{6x}}{e^{6x}+9}\,\, dx

    Let u = e^{6x} + 9\,\,\therefore du = 6e^{6x}\,\,dx

    Now plugging in u and du we get:

    \begin{align*}\int \frac{e^{6x}}{e^{6x}+9}\,\, dx=& \int \frac{1}{6u}\,\,du \\=& \frac{1}{6} \ln|u| + C\\=& \frac{1}{6}\ln| e^{6x}+9| + C.....\text{[by plugging in value of } u\text{]}\end{align*}

    \int \frac{e^{6x}}{e^{6x}+9}\,\, dx = \frac{1}{6}\ln| e^{6x}+9| + C.....\text{[Answer]}

    You can check the answer here:

    int (e^(6x))/(e^(6x)+9) - Wolfram|Alpha
    And of course, since \displaystyle \begin{align*} e^{6x} > 0 \end{align*} for all x, so is \displaystyle \begin{align*} e^{6x} + 9 \end{align*}, so the absolute value signs here are overkill
    Thanks from x3bnm
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