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Math Help - Integrate Using Substitution

  1. #1
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    Integrate Using Substitution

    I got this question in my maths exam today, I think I got it right - but I'm unsure.

    <br />
\int_0^2 \frac{2^x}{(2^x+1)^2} dx<br />

    Use the substitution u = 2^x

    I got something like 1/ln64.
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  2. #2
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    Quote Originally Posted by Killswitch View Post
    I got this question in my maths exam today, I think I got it right - but I'm unsure.

    <br />
\int_0^2 \frac{2^x}{(2^x+1)^2} dx<br />

    Use the substitution u = 2^x

    I got something like 1/ln64.
    \int_0^2 \frac{e^{x\ln 2}}{\left( e^{x\ln 2} + 1 \right)^2} \ dx

    Let t = x\ln 2

    \frac{1}{\ln 2} \int_0^{2\ln 2} \frac{e^t}{(e^t+1)^2} dt

    Let \chi = e^t+1

    \int_2^3 \frac{1}{\chi ^2} d \chi
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  3. #3
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    Yes, that looks like the right substitution... but I get an answer of \frac{3}{10\ln 2}.


    We have to evaluate \int_0^2 \frac{2^x}{(2^x+1)^2} dx :

    Substitute u = 2^x

    \Rightarrow \ln u = \ln 2^x = x \ln 2

    \Rightarrow \frac{1}{u} \frac{du}{dx} = \ln 2, i.e. u dx = \frac{du}{\ln 2}.


    So, \int_0^2 \frac{2^x}{(2^x+1)^2} dx = \int_0^2 \frac{u dx}{(u+1)^2} = \frac{1}{\ln 2} \int_1^4 \frac{du}{(u + 1)^2}.

    = \frac{1}{\ln 2} \left[ \frac{-1}{(u+1)}\right]_1^4 = \frac{1}{\ln 2} \cdot \left( - \frac{1}{5} + \frac{1}{2}\right) = \frac{3}{10\ln 2}.
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  4. #4
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    Here is another way.
    u = 2^x  + 1\quad  \Rightarrow \quad du = 2^x \ln (2)dx
    \int\limits_0^2 {\frac{{2^x dx}}{{\left( {2^x  + 1} \right)^2 }} = \frac{1}{{\ln (2)}}\int\limits_2^5 {u^{ - 2} du} } <br />
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  5. #5
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    Yet another way without changing the upper and lower limits
    Attached Thumbnails Attached Thumbnails Integrate Using Substitution-jun.gif  
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