Question

Помогите, пожалуйста!!!

Answer 1

Ответ:

$\int ^{ln2}_{0} \sqrt{ {e}^{x} - 1 } dx = \\ = | \:t = {e}^{x} ;dt = {e}^{x} dx ;{e}^{0} = 1 ;{e}^{ln2} = 2 | = \\ = \int^{2}_{1} \frac{ \sqrt{t - 1} }{t} dt = \\ = |s= \sqrt{t - 1};ds = \frac{dt}{2 \sqrt{t - 1} }; \sqrt{2 - 1} = 1; \sqrt{1 - 1} = 0| = \\ = 2\int^{1}_{0} \frac{ {s}^{2} }{ {s}^{2} + 1} ds = 2\int^{1}_{0} \frac{ {s}^{2} + 1 - 1}{ {s}^{2} + 1 } ds = \\ = 2\int^{1}_{0}(1 - \frac{1}{ {s}^{2} + 1 } )ds = 2(s - arctgs) |^{1}_{0} = 2 - \frac{\pi}{2}$