Question

Помогите пожалуйста ​

Answer 1

Ответ:

1.

$\int\limits \sin(3x + 5) dx = \frac{1}{3} \int\limits \sin(3x + 5)d(3x) = \\ = \int\limits\sin(3x + 5) d(3x + 5) = - \cos(3x + 5) + C$

2.

$\int\limits {e}^{2x} dx = \frac{1}{2}\int\limits {e}^{2x} d(2x) = {e}^{2x} + C$

3.

$\int\limits \: tgxdx = \int\limits \frac{ \sin(x) }{ \cos(x) } dx = - \int\limits \frac{d ( \cos(x) ) }{ \cos(x) } = \\ = - ln( | \cos(x) | ) + C$

4.

$\int\limits {e}^{ - {x}^{2} } xdx \\ \\ - {x}^{2} = t \\ - 2xdx = dt \\ xdx = - \frac{dt}{2} \\ \\ - \frac{1}{2} \int\limits {e}^{t} dt = - \frac{1}{2} {e}^{t} + C= - \frac{1}{2} {e}^{ - {x}^{2} } + C$

5.

$\int\limits \frac{e {}^{4x} }{ {e}^{x} - 1} dx = \int\limits \frac{ {e}^{4x} - 1 + 1}{ {e}^{x} - 1}dx = \\ = \int\limits \frac{ {e}^{4x} - 1}{ {e}^{x} - 1 }dx + \int\limits \frac{ 1 }{ {e}^{x} - 1} dx = \\ = \int\limits \frac{ ({e}^{2x} - 1)( {e}^{2x} + 1) }{ {e}^{x} - 1} dx + \int\limits \frac{dx}{e {}^{x} - 1} = \\ = \int\limits \frac{( {e}^{x} - 1)( {e}^{x} + 1)( {e}^{2x} + 1) }{ {e}^{x} - 1 } + \int\limits \frac{dx}{e {}^{x} - 1} = \\ = \int\limits( {e}^{x} + 1)( {e}^{2x} + 1) dx+\int\limits \frac{dx}{e {}^{x} - 1 } \\ \\ 1)\int\limits( {e}^{3x} + {e}^{x} + {e}^{2x} + 1)dx = \\ = \frac{1}{3} \int\limits e {}^{3x} d(3x) + \int\limits {e}^{x} dx + \frac{1}{2} \int\limits {e}^{2x} d(2x) + x = \\ = \frac{1}{3} {e}^{3x} + {e}^{x} + \frac{1}{2} {e}^{2x} + x + c \\ \\ 2)\int\limits \frac{dx}{e {}^{x} - 1} \\ \\ {e}^{x} - 1 = t \\ e {}^{x} dx = dt \\ dx = \frac{dt}{t} \\ \\ \int\limits \frac{dt}{t(t - 1)} =- \int\limits \frac{dt}{t - 1} + \int\limits \frac{dt}{t} = \\ = - ln( |t - 1| ) + ln( |t| ) + C = \\ = - ln( {e}^{x} ) + ln( | {e}^{x} - 1 | ) + C = \\ = - x + ln( | {e}^{x} - 1 | ) + C$

получаем:

$\frac{1}{3} {e}^{3x} + \frac{1}{2} {e}^{2x} + {e}^{x} + x - x + ln( | {e}^{x} - 1 | ) + C= \\ = \frac{ {e}^{3x} }{3} + \frac{ {e}^{2x} }{2} + {e}^{x} + ln( | {e}^{x} - 1 | ) + C$

6.

$\int\limits \frac{ {x}^{4} }{ {x}^{5} + 7 } dx = \frac{1}{5} \int\limits \frac{5 {x}^{4} }{ {x}^{5} + 7} dx = \\ = \frac{1}{5} \int\limits \frac{d( {x}^{5} + 7)}{ {x}^{5} + 7} = \frac{1}{5} ln( | {x}^{5} + 7 | ) + C$

7.

$\int\limits \frac{dx}{ \cos {}^{2} (3x) } = \frac{1}{3} \int\limits \frac{d(3x)}{ \cos {}^{2} (3x) } = \frac{1}{3} tg(3x) + C$

8.

$\int\limits \frac{ \sqrt{x} }{ \sqrt[3]{x} + 1 } dx \\ \\ {x}^{ \frac{1}{6} } = t \\ \sqrt{x} = {t}^{3} \\ \sqrt[3]{x} = {t}^{2} \\ x = {t}^{6} \\ dx =6 t {}^{5} dt \\ \int\limits \frac{ {t}^{3} \times 6 {t}^{5} }{ {t}^{2} + 1} dt = 6\int\limits \frac{ {t}^{8} }{t {}^{2} + 1} dt = \\ = 6( {t}^{6} - t {}^{4} + {t}^{2} - 1 + \frac{1}{ {t}^{2} + 1 } )dt = \\ = 6( \frac{ {t}^{7} }{7} - \frac{ {t}^{5} }{5} + \frac{ {t}^{3} }{3} - t + arctg(t)) + C = \\ = \frac{6}{7} x \sqrt[6]{ {x}^{} } - \frac{6}{5} \sqrt[6]{ {x}^{5} } + 2 \sqrt{x} - 6\sqrt[6]{x} + 6arctg( \sqrt[6]{x} ) + C$

9.

$\int\limits \frac{ \sqrt{x + 1} + 1}{ \sqrt{x + 1} - 1 } dx = \int\limits \frac{ \sqrt{x + 1} - 1 + 2 }{ \sqrt{x + 1} - 1 } dx = \\ = \int\limits(1 + \frac{2}{ \sqrt{x + 1} - 1} )dx \\ \\ \int\limits \frac{2dx}{ \sqrt{x + 1} - 1} \\ \\ \sqrt{x + 1 } - 1 = t \\ \frac{dx}{2 \sqrt{x + 1} }dx = dt \\ dx = 2 \sqrt{x + 1}d t \\ dx = 2(t +1)dt \\ \\ 2\int\limits \frac{2(t + 1)}{t} dt = 4\int\limits(1 + \frac{1}{t} )dt = 4t +4 ln( |t| ) + C= \\ = 4( \sqrt{x + 1} - 1) + 4( \sqrt{x + 1} - 1) + C\\ \\ - - - - - - - - - - - - \\ = x + 4 ln( | \sqrt{x + 1} - 1 | ) + 4( \sqrt{x + 1} - 1) + C$

10.

$\int\limits \frac{dx}{x (ln(x) + 1) } \\ \\ ln(x) + 1 = t \\ \frac{dx}{x} = dt \\ \\ \int\limits \frac{dt}{t} = ln( |t| ) + C = ln( | ln(x) + 1| ) + C$

11.

$\int\limits {e}^{ \cos(x) } \sin(x)dx = - \int\limits {e}^{ \cos(x) }( - \sin(x)) dx = \\ = - \int\limits {e}^{ \cos(x) } d( \cos(x)) = - {e}^{ \cos(x) } + C$

12.

$\int\limits \frac{ \sqrt{1 + ln(x) } }{x} dx = \int\limits {(1 + ln(x)) }^{ \frac{1}{2} } d (ln(x)) = \\ = \int\limits(1 + ln(x)) {}^{ \frac{1}{2} } d( ln(x) + 1) = \frac{ {(1 + ln(x)) }^{ \frac{3}{2} } }{ \frac{3}{2} } + C = \\ = \frac{2}{3} \sqrt{ {(1 + ln(x)) }^{3} } + C$