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 4 - 2007 - Paper 8

The integral (\int \frac{1}{(u + 1)^2} , du) evaluates to (-\frac{1}{u + 1} + C). Therefore:

[ \frac{1}{\ln(2)} \left(-\frac{1}{u + 1}\right) + C. ]

When changing the limits of integration, evaluate at the original limits:

• When (x = 0), (u = 2^0 = 1).
• When (x = 1), (u = 2^1 = 2). Thus, we evaluate the integral from (1) to (2).

We now calculate:

[ = \frac{1}{\ln(2)} \left( -\frac{1}{2} + \frac{1}{1} \right) = \frac{1}{\ln(2)} \left( 1 - \frac{1}{2} \right) = \frac{1}{2 \ln(2)}. ] Therefore, the exact value is (\frac{1}{2 \ln(2)}).