`int_0^ln2 2e^(-x)coshx dx` Evaluate the integral

To compute this integral, recall the definition of hyperbolic cosine function:  `cosh(x) = (e^x + e^(-x))/2.`

Therefore the function under integral is equal to `1 + e^(-2x),` and its indefinite integral is  `x - 1/2 e^(-2x) + C.`

Now we can substitute the given limits of integration to the antiderivative and obtain

`int_0^(ln2) 2e^(-x) cosh(x) dx = ln2 - 1/2 (e^(-2ln2)) - (0 - 1/2) =`

`= ln2 - 1/8 + 1/2 =` ln2 + 3/8

(we used the identity `e^(lny) = y` and the property of exponent  `(a^b)^c = a^(bc)` ).