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Math Help - integrals involving multiple e's

  1. #1
    Junior Member TheMathTham's Avatar
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    integrals involving multiple e's

    The problem is: \int \frac{e^{3x}}{e^{6x}+1}

    By using multiple product rules, I ended up with \frac{5^{2x}}{x}+2ln(5)ln(3x)

    The answer choice I selected had the same answer, except that it had 5^{2x} at the end. \frac{5^{2x}}{x}+2ln(5)ln(3x)5^{2x}

    Which one is correct? If it is the second one, can you please explain where the additional 5^{2x} came from?
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  2. #2
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    Quote Originally Posted by TheMathTham View Post
    The problem is: \int \frac{e^{3x}}{e^{6x}+1}

    By using multiple product rules, I ended up with \frac{5^{2x}}{x}+2ln(5)ln(3x)

    The answer choice I selected had the same answer, except that it had 5^{2x} at the end. \frac{5^{2x}}{x}+2ln(5)ln(3x)5^{2x}

    Which one is correct? If it is the second one, can you please explain where the additional 5^{2x} came from?
    Use u=e^{3x} and then remember the arctan
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  3. #3
    MHF Contributor Calculus26's Avatar
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    e^(6x) = [e^(3x)]^2

    Let u = e^(3x)

    you should obtain 1/3 arctan(e^3x) + C

    what multiple product rules do you use on an integral?
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  4. #4
    Junior Member TheMathTham's Avatar
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    oh wow i just realized i wrote the problem for one question and then the work to a different question that i was having trouble with

    hehe ok so fail there by me...

    The problem to the work was \frac{\mathrm{d} }{\mathrm{d} x}(ln(3x))5^{2x}). If you could help on this one too that would be great

    And as for the original problem, I understand the answer now. But I'm curious, why is the \int \frac{u}{u^{2}+1}=\arctan (u) ?

    Sorry for the mess up
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  5. #5
    MHF Contributor Calculus26's Avatar
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    The second question first

    for integral [ e^(3x)/(e^(6x)+1)dx]

    if u = e^(3x) du = 3 e^(3x) dx

    so you get 1/3 integral [1/(u^2 +1)du] which is the arctan(u)

    For the first

    [a^(u)] ' = ln(a) a^(u) du/dx

    [5^(2x)] ' = ln(5)5^(2x)2
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  6. #6
    Junior Member TheMathTham's Avatar
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    Quote Originally Posted by Calculus26 View Post
    The second question first

    for integral [ e^(3x)/(e^(6x)+1)dx]

    if u = e^(3x) du = 3 e^(3x) dx

    so you get 1/3 integral [1/(u^2 +1)du] which is the arctan(u)

    For the first

    [a^(u)] ' = ln(a) a^(u) du/dx

    [5^(2x)] ' = ln(5)5^(2x)2

    For the second question, I was asking why the \int \frac{u}{u^{2}+1}=\arccos (u). I thought that  \frac{u}{u^{2}+1}=\arccos (u) not the  \int \frac{u}{u^{2}+1}=\arccos (u)

    For the first question...I'm not following how you arrived there. What I did was to use the product rule and make u=ln(3x), u{}'=\frac{1}{x}, v=5^{2x}, v{}'=2ln(5). I arrived at v{}' by substituting 2xln(5) in for 5^{2x} and then using the product rule again in which the new variables were u=2x, u{}'=2, v=ln(5), v{}'=0. The final result was \frac{5^{2x}}{x}+2ln(5)ln(3x). Is that correct?
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  7. #7
    MHF Contributor Calculus26's Avatar
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    for 5^(2x)

    [5^(2x)] ' = ln(5)*5^(2x)*2

    as in general [a^u]' = ln(a) *a^u du/dx

    Eg (2^(x)) ' = ln(2) 2^x

    (2^(sin(x)) ' = ln(2)2^(sin(x)) cos(x)

    By the way integral ( u/(u^2+1)du) = 1/2 ln(u^2+1)
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  8. #8
    Junior Member TheMathTham's Avatar
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    Quote Originally Posted by Calculus26 View Post
    for 5^(2x)

    [5^(2x)] ' = ln(5)*5^(2x)*2

    as in general [a^u]' = ln(a) *a^u du/dx

    Eg (2^(x)) ' = ln(2) 2^x

    (2^(sin(x)) ' = ln(2)2^(sin(x)) cos(x)

    By the way integral ( u/(u^2+1)du) = 1/2 ln(u^2+1)
    ohhhhhh i get it. thanks a ton!
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