Алгебра: решение интегральных задачи

решение интегральных задачи

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Ответ:

Jamilya28

19.04.2021 13:10

$\int\limits \frac{dx}{x \sqrt{ {x}^{2} - 2} } \\ \\ \sqrt{ {x}^{2} - 2} = t \\ {x}^{2} - 2 = {t}^{2} \\ 2xdx = 2tdt \\ xdx = tdt \\ dx = \frac{tdt}{x} \\ \\ \int\limits \frac{tdt}{x} \times \frac{1}{x \times t} = \int\limits \frac{dt}{ {x}^{2} } \\ \\ {x}^{2} = 2 + {t}^{2} \\ \\ \int\limits \frac{dt}{t {}^{2} + 2} = \int\limits \frac{dt}{t {}^{2} + {( \sqrt{2}) }^{2} } = \\ = \frac{1}{ \sqrt{2} } arctg( \frac{t}{ \sqrt{2} } )+ C = \\ = \frac{1}{ \sqrt{2} } arctg( \frac{ \sqrt{ {x}^{2} - 2 } }{ \sqrt{2} } ) + C$

$\int\limits \frac{1 + ln(x) }{x}dx \\ \\ 1 + ln(x) = t \\ \frac{dx}{x} = dt \\ \\ \int\limits \: tdt = \frac{ {t}^{2} }{2} + C = \frac{ {(1 + ln(x)) }^{2} }{2} + C$

$\int\limits \frac{dx}{e {}^{x} + 1 } \\ \\ {e}^{x} + 1 = t \\ {e}^{x} = t - 1\\ e {}^{x} dx = dt \\ dx = \frac{dt}{e {}^{ x} } = \frac{dt}{t - 1} \\ \\ \int\limits \frac{dt}{t(t - 1)} \\ \\ \frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \\ 1 = A( t- 1) + Bt \\1 = At - A + B t \\ \\$

$\left \{ {{0 = A + B} \atop { 1 = - A} } \right. \\ \\ \left \{ {{A = - 1} \atop {B= 1} } \right.$

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

$\int\limits \frac{xdx}{ \sqrt{x + 1} } \\ \\ \sqrt{x + 1} = t \\ x + 1 = {t}^{2} \\ x = {t}^{2} - 1 \\ dx = 2tdt \\ \\ \int\limits \frac{( {t}^{2} - 1) \times 2tdt }{t} = 2\int\limits( {t}^{2} - 1) dt = \\ = 2( \frac{ {t}^{3} }{3} - t) + C= \frac{2 \sqrt{ {(x + 1)}^{3} } }{3} - 2 \sqrt{x +1 } + C$

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

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

$\int\limits \frac{x}{ {(3x + 2)}^{3} } dx = \frac{1}{3} \int\limits \frac{3xdx}{ {(3x + 2)}^{3} } = \\ = \frac{1}{3} \int\limits \frac{3x + 2 - 2}{ {(3x + 2)}^{3} } dx = \\ = \frac{1}{3} \int\limits \frac{dx}{ {(3x + 2)}^{2} } - \frac{2}{3} \int\limits \frac{dx}{ {(3x + 2)}^{3} } = \\ = \frac{1}{9} \int\limits \frac{d(3x)}{ {(3x + 2)}^{2} } - \frac{2}{9} \int\limits \frac{d(3x)}{ {(3x + 2)}^{3} } = \\ = \frac{1}{9} \int\limits \frac{d(3x + 2)}{ {(3x + 2)}^{2} } - \frac{2}{9} \int\limits \frac{d(3x + 2)}{ {(3x + 2)}^{3} } = \\ = \frac{1}{9} \times \frac{ {(3x + 2)}^{ - 1} }{( - 1)} - \frac{2}{9} \times \frac{ {(3x + 2)}^{ - 2} }{( - 2)} + C = \\ = - \frac{1}{9 (3x + 2)} + \frac{1}{9 {(3x + 2)}^{2} } + C= \\ = \frac{1}{9(3x + 2)} ( \frac{1}{3x + 2} - 1) + C = \\ = \frac{1}{9(3x + 2)} \times \frac{1 - 3x -2 }{3x + 2} + C = \\ = \frac{ - 1 - 3x}{9 {(3x + 2)}^{2} } + C = - \frac{1 + 3x}{9 {(3x + 2)}^{2} } + C$