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Math Help - solving an integral using the table

  1. #1
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    solving an integral using the table

    Hi,

    Could someone please tell me if this is correct?

    int. e^2x/sq.2+e^x

    int. du/a^2+b^2=1/a arctanu/a+c
    u=2+e^x
    du=e^xdx

    int. e^2x/sq. 2+e^x=1/sq.2 arctan 2+e^x/sq. 2+c

    Thank you
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  2. #2
    MHF Contributor red_dog's Avatar
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    It's about \displaystyle\int\frac{e^{2x}}{2^2+e^x}dx or \displaystyle\int\frac{e^{2x}}{\sqrt{2+e^x}}dx ?
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  3. #3
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    I think it's the second one.

    Just set u=\sqrt{2+e^x}
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  4. #4
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    This one

    Sorry for the confusion
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  5. #5
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    Just set
    Do I just plug this into the forumula or do I need to find du?

    Is du 1/2(2+e^x)^-1/2???

    Thank you
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  6. #6
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    Quote Originally Posted by chocolatelover View Post
    Is du 1/2(2+e^x)^-1/2???
    You forgot the chain rule, try again
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  7. #7
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    Is du 1/2(2+e^x)^-1/2???
    You forgot the chain rule
    OOPs!

    so du is 1/2(2+e^x)^-1/2(e^x). This won't work for u substitution, right? It wouldn't just be 1/sq. 2art sq. 2+e^x/sq. 2+e^x+c, would it?

    Thank you
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  8. #8
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    \int\frac{e^{2x}}{\sqrt{2+e^x}}\,dx

    If u=\sqrt{2+e^x}\implies du=\frac{e^x}{2\sqrt{2+e^x}}\,dx, the integral becomes to

    \int\frac{e^{2x}}{\sqrt{2+e^x}}\,dx=2\int(u^2-2)\,du

    That's easy, don't you?
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  9. #9
    MHF Contributor red_dog's Avatar
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    u=\sqrt{2+e^x}\Rightarrow u^2=2+e^x\Rightarrow 2udu=e^xdx
    Substitute and you'll get \displaystyle2\int(u^2-2)du
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hi,

    Could someone please tell me if this is correct?

    int. e^2x/sq.2+e^x

    int. du/a^2+b^2=1/a arctanu/a+c
    u=2+e^x
    du=e^xdx

    int. e^2x/sq. 2+e^x=1/sq.2 arctan 2+e^x/sq. 2+c

    Thank you
    Please do us a favor and either learn LaTeX or start using parenthesis! The way you are writing this is confusing.

    -Dan
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