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    I have been given a question and I have tried multiple methods and I still can't get the answer of '1/6ln2' or '1/ln4 - 1/ln8'.

    Without further ado,

    Use the substitution u=2^x to find the exact value of

    2^x/(2^x+1)^2 dx. Limits of x are 1 and 0.

    I have tried all kinds of things, and I'd very much appreciate some form of way doing this.
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    (Original post by JLangley753)
    I have been given a question and I have tried multiple methods and I still can't get the answer of '1/6ln2' or '1/ln4 - 1/ln8'.

    Without further ado,

    Use the substitution u=2^x to find the exact value of

    2^x/(2^x+1)^2 dx. Limits of x are 1 and 0.

    I have tried all kinds of things, and I'd very much appreciate some form of way doing this.
    Hint:

    \dfrac{1}{\ln 2} (\ln 2) = 1

    Use this fact when making the substitution.
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    (Original post by Indeterminate)
    Hint:

    \dfrac{1}{\ln 2} (\ln 2) = 1

    Use this fact when making the substitution.
    I'm at the stage where I've got 1/(u+1)^2 du with limits 2 and 1.

    Is this where I'd input this into the substitution?
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    (Original post by JLangley753)
    I'm at the stage where I've got 1/(u+1)^2 du with limits 2 and 1.

    Is this where I'd input this into the substitution?
    You get

    \dfrac{du}{dx} = 2^x \ln 2 \Rightarrow du = 2^x \ln 2 \ dx

    Now, you only have a 2 to the x on the top. So if you multiply this by ln 2 to substitute the above, you need to divide by ln 2 too to AVOID changing the integral.
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    (Original post by Indeterminate)
    You get

    \dfrac{du}{dx} = 2^x \ln 2 \Rightarrow du = 2^x \ln 2 \ dx

    Now, you only have a 2 to the x on the top. So if you multiply this by ln 2 to substitute the above, you need to divide by ln 2 too to AVOID changing the integral.
    Surely then I'm left with the equivalent u^2/(u+1)^2 but with ln2 on the top and bottom?

    I hope this comes to me eventually.
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    (Original post by JLangley753)
    Surely then I'm left with the equivalent u^2/(u+1)^2 but with ln2 on the top and bottom?

    I hope this comes to me eventually.
    No.

    du = 2^x \ln 2 \ dx

    You want to replace the RHS with the LHS in the integral, right?

    Therefore, you have to multiply it ln 2 and then that numerator all becomes du.

    Now, you've changed the integral by multiplying by ln 2, so you have to divide the integral that's with respect to u i.e the one with du in it, by ln 2 as well
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    (Original post by JLangley753)
    I have been given a question and I have tried multiple methods and I still can't get the answer of '1/6ln2' or '1/ln4 - 1/ln8'.

    Without further ado,

    Use the substitution u=2^x to find the exact value of

    2^x/(2^x+1)^2 dx. Limits of x are 1 and 0.

    I have tried all kinds of things, and I'd very much appreciate some form of way doing this.

    \displaystyle \int \dfrac{2^x}{(2^x+1)^2} dx

    u = 2^x

    gives

    \dfrac{du}{dx} = u \ln 2

    So  dx = \dfrac{du}{u \ln 2}

    so

    \displaystyle \int \dfrac{2^x}{(2^x+1)^2} dx = \displaystyle \int \dfrac{u}{(u+1)^2} \times \dfrac{1}{u \ln 2} du
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    (Original post by TenOfThem)
    \displaystyle \int \dfrac{2^x}{(2^x+1)^2} dx

    u = 2^x

    gives

    \dfrac{du}{dx} = u \ln 2

    So  dx = \dfrac{du}{u \ln 2}

    so

    \displaystyle \int \dfrac{2^x}{(2^x+1)^2} dx = \displaystyle \int \dfrac{u}{(u+1)^2} \times \dfrac{1}{u \ln 2} du
    I've got this, absolutely fine. I know the integral of the first part is 1/u+1 + ln(u+1) but do I need to integrate 1/uln2? I have never integrated something like that before.
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    1/ln2 is a constant...

    EDIT: and where does integrating 1/uln2 come from? You only need to integrate 1/(u+1)^2 (times 1/ln2)
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    (Original post by JLangley753)
    I've got this, absolutely fine. I know the integral of the first part is 1/u+1 + ln(u+1) but do I need to integrate 1/uln2? I have never integrated something like that before.
    erm, no

    you need to cancel the u in the numerator with the u in the denominator

    then integrate
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    Because you say "integral of the first part", perhaps your error is that you're treating the multiplication sign as an addition sign?
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    (Original post by TenOfThem)
    erm, no

    you need to cancel the u in the numerator with the u in the denominator

    then integrate
    Argh, thank you. Sorry, I actually don't know how I didn't get this before. I should have realised what Indeterminate was talking about in his first post.

    From here I took the ln2 outside, integrated to get -(u+1)^-1 and the rest is a treat. Input the limits to get 1/6, multiply that by ln2 to get 1/6ln2.

    Thank you guys, seriously. I apologise for being so slow at times.
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    (Original post by JLangley753)
    Argh, thank you. Sorry, I actually don't know how I didn't get this before. I should have realised what Indeterminate was talking about in his first post.

    From here I took the ln2 outside, integrated to get -(u+1)^-1 and the rest is a treat. Input the limits to get 1/6, multiply that by ln2 to get 1/6ln2.

    Thank you guys, seriously. I apologise for being so slow at times.
 
 
 
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