11. \( \int_{0}^{2} \frac{e^{x}}{1+e^{2 x}} d x \)

To evaluate the integral \( \int_{0}^{2} \frac{e^{x}}{1+e^{2 x}} d x \), we can use substitution to simplify our calculations. First, notice that \( e^{2x} = (e^x)^2 \). Let's set \( u = e^x \), then \( du = e^x \, dx \) or \( dx = \frac{du}{u} \). The limits change accordingly; when \( x = 0 \), \( u = e^0 = 1 \), and when \( x = 2 \), \( u = e^2 \). Substituting, we have: \[ \int_{1}^{e^2} \frac{u}{1 + u^2} \cdot \frac{du}{u} = \int_{1}^{e^2} \frac{1}{1 + u^2} \, du \] This integral simplifies to: \[ \int_{1}^{e^2} \frac{1}{1 + u^2} \, du \] The integral of \( \frac{1}{1 + u^2} \) is \( \tan^{-1}(u) \). Therefore, we have: \[ \tan^{-1}(u) \bigg|_{1}^{e^2} = \tan^{-1}(e^2) - \tan^{-1}(1) \] Since \( \tan^{-1}(1) = \frac{\pi}{4} \), the final result will be: \[ \tan^{-1}(e^2) - \frac{\pi}{4} \] Thus, the value of the integral is: \[ \int_{0}^{2} \frac{e^{x}}{1+e^{2 x}} d x = \tan^{-1}(e^2) - \frac{\pi}{4} \]