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.
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, $\displaystyle \int_0^2 \frac{2^x}{(2^x+1)^2} dx = \int_0^2 \frac{u dx}{(u+1)^2} = \frac{1}{\ln 2} \int_1^4 \frac{du}{(u + 1)^2}$.
$\displaystyle = \frac{1}{\ln 2} \left[ \frac{-1}{(u+1)}\right]_1^4 = \frac{1}{\ln 2} \cdot \left( - \frac{1}{5} + \frac{1}{2}\right) = \frac{3}{10\ln 2}$.