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Math Help - Another Integral

  1. #1
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    Another Integral

    Evaluate the indefinite integral. (Remember to use ln(abs(u)) where appropriate.)

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    Quote Originally Posted by Jgirl689 View Post
    Evaluate the indefinite integral. (Remember to use ln(abs(u)) where appropriate.)

    \int{\frac{e^{9x}}{e^{9x} + 3}\,dx} = \frac{1}{9}\int{\frac{9e^{9x}}{e^{9x} + 3}\,dx}.

    Now make the substitution u = e^{9x} in the denominator.
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  3. #3
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    u = e^9x
    (1/9) du = e^9x dx
    int((1 / u + 3)du)
    ln|u + 3|
    ln|e^9x + 3| + C
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    Quote Originally Posted by Locke View Post
    u = e^9x
    (1/9) du = e^9x dx
    int((1 / u + 3)du)
    ln|u + 3|
    ln|e^9x + 3| + C
    Correct.

    But since e^{9x} > 0 for all x, you don't need the absolute value.
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  5. #5
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    I put that in my answer thing, and it was marked wrong, should I use log instead?
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