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Thread: 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.

    $\displaystyle
    \int_0^2 \frac{2^x}{(2^x+1)^2} dx
    $

    Use the substitution $\displaystyle u = 2^x$

    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.

    $\displaystyle
    \int_0^2 \frac{2^x}{(2^x+1)^2} dx
    $

    Use the substitution $\displaystyle u = 2^x$

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

    Let $\displaystyle t = x\ln 2$

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

    Let $\displaystyle \chi = e^t+1$

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


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

    Substitute $\displaystyle u = 2^x$

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

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


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

    $\displaystyle = \frac{1}{\ln 2} \left[ \frac{-1}{(u+1)}\right]_1^4 = \frac{1}{\ln 2} \cdot \left( - \frac{1}{5} + \frac{1}{2}\right) = \frac{3}{10\ln 2}$.
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  4. #4
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    Here is another way.
    $\displaystyle u = 2^x + 1\quad \Rightarrow \quad du = 2^x \ln (2)dx$
    $\displaystyle \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} }
    $
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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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