Помогите пожалуйста!??. Created by symbatamzebek40. algebra-ru

Помогите пожалуйста!??...

symbatamzebek40 @symbatamzebek40

September 2021 1 32 Report

Помогите пожалуйста!??

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Miroslava227

Ответ:

$\int\limits^{1}_{0} \frac{ \sqrt{x} }{ \sqrt{x} + 1 } dx \\ \\ \sqrt{x} + 1 = t \\ \sqrt{x} = t - 1 \\ \frac{1}{2 \sqrt{x} } dx = dt \\ dx = 2 \sqrt{x} dt\\ dx = 2(t - 1)dt \\ t1 = \sqrt{1} + 1 = 2 \\ t2 = \sqrt{0} + 1 = 1 \\ \\ \int\limits^{ 2 } _ {1} \frac{t - 1}{t} \times 2(t - 1)dt = \\ =2 \int\limits^{ 2 } _ {1} \frac{ {t}^{2} - 2t + 1}{t}dt = 2 \int\limits^{ 2 } _ {1}(t - 2 + \frac{1}{ t} )dt = \\ = 2( \frac{ {t}^{2} }{2} - 2t + ln( |t| ) )| ^{ 2 } _ {1} = ( {t}^{2} - 4t + 2 ln( |t| )) | ^{ 2 } _ {1} = \\ = 4 - 8 + 2 ln(2) - 1 + 4 - 2 ln(1) = \\ = 2 ln(2) - 1$

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