Evaluate the definite integral of (e^x + 2e^-x)2 from 0 to ln 2.

We use the formula $(a+b)^2 = a^2 + 2ab + b^2$.

$$(e^x + 2e^{-x})^2 = (e^x)^2 + 2(e^x)(2e^{-x}) + (2e^{-x})^2$$ $$= e^{2x} + 4e^{x-x} + 4e^{-2x}$$ $$= e^{2x} + 4e^0 + 4e^{-2x}$$ $$= e^{2x} + 4 + 4e^{-2x}$$

We integrate term by term:

$$\int (e^{2x} + 4 + 4e^{-2x})\,dx = \int e^{2x}\,dx + \int 4\,dx + \int 4e^{-2x}\,dx$$ $$= \frac{1}{2}e^{2x} + 4x + 4\left(-\frac{1}{2}\right)e^{-2x}$$ $$= \frac{1}{2}e^{2x} + 4x - 2e^{-2x}$$

Now we evaluate the definite integral from 0 to $\ln 2$:

$$\left[ \frac{1}{2}e^{2x} + 4x - 2e^{-2x} \right]_{0}^{\ln 2}$$

$$\frac{1}{2}e^{2\ln 2} + 4\ln 2 - 2e^{-2\ln 2}$$

Using the logarithm property $a\ln b = \ln(b^a)$ and $e^{\ln k} = k$:

$$= \frac{1}{2}e^{\ln(2^2)} + 4\ln 2 - 2e^{\ln(2^{-2})}$$ $$= \frac{1}{2}e^{\ln 4} + 4\ln 2 - 2e^{\ln(1/4)}$$ $$= \frac{1}{2}(4) + 4\ln 2 - 2\left(\frac{1}{4}\right)$$ $$= 2 + 4\ln 2 - \frac{1}{2}$$ $$= \frac{4}{2} - \frac{1}{2} + 4\ln 2 = \frac{3}{2} + 4\ln 2$$

$$\frac{1}{2}e^{2(0)} + 4(0) - 2e^{-2(0)}$$ $$= \frac{1}{2}e^0 + 0 - 2e^0$$ $$= \frac{1}{2}(1) - 2(1)$$ $$= \frac{1}{2} - 2 = -\frac{3}{2}$$

$$\left( \frac{3}{2} + 4\ln 2 \right) - \left( -\frac{3}{2} \right)$$ $$= \frac{3}{2} + 4\ln 2 + \frac{3}{2}$$ $$= \frac{6}{2} + 4\ln 2$$ $$= 3 + 4\ln 2$$

The final answer is $3 + 4\ln 2$.