Use the substitution $u = 2^x$ to find the exact value of $$\int_0^1 \frac{2^x}{(2^x + 1)^2} \, dx.$$ - Edexcel - A-Level Maths Pure - Question 3 - 2007 - Paper 8

After substituting, the integral becomes:

$\int_1^2 \frac{u}{(u + 1)^2} \cdot \frac{du}{u \ln(2)} = \frac{1}{\ln(2)} \int_1^2 \frac{1}{(u + 1)^2} \, du.$

Next, we compute the integral:

$\int \frac{1}{(u + 1)^2} \, du = -\frac{1}{u + 1} + C.$

Evaluating this from $u=1$ to $u=2$ gives:

$\left[-\frac{1}{u + 1}\right]_1^2 = -\frac{1}{3} - (-\frac{1}{2}) = -\frac{1}{3} + \frac{1}{2} = \frac{1}{6}.$

Therefore, the integral yields:

$\frac{1}{\ln(2)} \cdot \frac{1}{6} = \frac{1}{6\ln(2)}.$

Combining these results, the exact value of the integral is:

$\frac{1}{6 \ln(2)}.$