Question

Помогите пожалуйста ​

Answer 1

Ответ:

17.

$\int\limits \frac{dx}{ \sqrt{ {e}^{x} + 1 } } \\ \\ \sqrt{ {e}^{x} + 1} = t \\ {e}^{x} + 1 = {t}^{2} \\ {e}^{x} = {t}^{2} - 1 \\ \frac{1}{ 2\sqrt{ {e}^{x} + 1 } } \times {e}^{x} dx = dt \\ dx = 2 \frac{ \sqrt{ {e}^{x} + 1 } }{ {e}^{x} } dt \\ dx = \frac{2t}{t - 1} dt \\ \\ \int\limits \frac{1}{t} \times \frac{2t}{t {}^{2} - 1} dt = 2\int\limits \frac{dt}{t {}^{2} - 1} = \\ = \frac{2}{2} ln( | \frac{t - 1}{t + 1} | ) + C = \\ = ln( | \frac{ \sqrt{{e}^{x} + 1} - 1}{ \sqrt{ {e}^{x} + 1 } + 1 } | ) + C$

18.

$\int\limits \frac{ {3}^{ \frac{1}{x} } }{ {x}^{2} } dx \\ \\ \frac{1}{x} = t \\ ( {x}^{ - 1} )'dx = dt \\ - {x}^{ - 2} dx = dt \\ \frac{dx}{ {x}^{2} } = - dt \\ \\ - \int\limits {3}^{t} dt = - \frac{ {3}^{t} }{ ln(3) } + C = - \frac{ {3}^{ \frac{1}{x} } }{ ln(3) } + C$

19.

$\int\limits \frac{ {(arctgx)}^{100} }{ {x}^{2} + 1 } dx \\ \\ arctgx = t \\ \frac{dx}{ {x}^{2} + 1 } = dt \\ \\ \int\limits {t}^{100} dt = \frac{ {t}^{101} }{101} + C = \frac{1}{101} {arctg}^{101} x + C$

20.

$\int\limits \frac{ {e}^{x} }{ \sqrt{4 - {e}^{2x} } } dx = \int\limits \frac{d( {e}^{x} )}{ \sqrt{ {2}^{2} - {(e {}^{x} )}^{2} } } = \\ = arcsin( \frac{ {e}^{x}}{2}) + C$

21.

$\int\limits \frac{ \cos( \sqrt{x} ) }{ \sqrt{x} } dx \\ \\ \sqrt{x} = t \\ \frac{1}{2 \sqrt{x} } dx = dt \\ \frac{dx}{ \sqrt{x} } = 2 dt \\ \\ 2\int\limits \cos(t) dt = 2 \sin(t) + C = \\ = 2\sin( \sqrt{x} ) + C$

22.

$\int\limits \frac{dx}{ {arccos}^{5}x \times \sqrt{1 - x {}^{2} } } dx \\ \\ arccosx = t \\ - \frac{1}{ \sqrt{1 - {x}^{2} } } dx = dt \\ \frac{1}{ \sqrt{1 - {x}^{2} } } dx = - dt \\ \\ - \int\limits \frac{dt}{t {}^{5} } = - \int\limits {t}^{ - 5} dt = \\ = - \frac{ {t}^{ - 4} }{( - 4)} + C = \frac{1}{4 {t}^{4} } + C= \\ = \frac{1}{4 {arccos}^{4} (x)} + C$