нужна .
Нужно решить методом замены неопределенного интеграла

нужна .Нужно решить методом замены неопределенного интеграла

Ответ:

LeraSmile12

11.04.2021 20:37

$\int\limits \frac{dx}{ \sin {}^{2} (1 - \frac{9x}{4} ) } \\ \\ 1 - \frac{9x}{4} = t \\ - \frac{9}{4} dx = dt \\ dx = - \frac{4}{9} dt \\ \\ - \frac{4}{9} \int\limits \frac{dt}{ \sin {}^{2} (t) } = - \frac{4}{9} \times ( - \ctg(t)) + C = \\ = \frac{4}{9} \ctg(1 - \frac{9x}{4} ) + C$

$\int\limits \frac{2 {x}^{4} + 7 {x}^{2} - 3 \sqrt{x} }{3 {x}^{3} } dx = \int\limits( \frac{2 {x}^{4} }{3 {x}^{3} } + \frac{7 {x}^{2} }{3 {x}^{3} } - \frac{3 \sqrt{x} }{3 {x}^{3} } )dx = \\ = \int\limits( \frac{2}{3} x + \frac{7}{3x} - {x}^{ - \frac{5}{2} } )dx = \frac{2}{3} \times \frac{ {x}^{2} }{2} + \frac{7}{3} ln |x| - \frac{ {x}^{ - \frac{3}{2} } }{( - \frac{3}{2} )} + C= \\ = \frac{ {x}^{2} }{3} + \frac{7}{3} ln |x| + \frac{2}{3x \sqrt{x} } + C$

$\int\limits \: tg( \frac{x}{9} + 55)dx = \int\limits \frac{ \sin( \frac{x}{9} + 55 ) }{ \cos( \frac{x}{9} + 55 ) } dx \\ \\ \cos( \frac{x}{9} + 55) = t \\ - \sin( \frac{x}{9} + 55) \times ( \frac{x}{9} + 55)' dx = dt \\ - \sin( \frac{x}{9} + 55 ) \times \frac{1}{9} dx = dt \\ \sin( \frac{x}{9} + 55) dx = - 9dt \\ \\ \int\limits \frac{( - 9)dt}{t} = - 9 ln |t| + C = \\ = - 9ln | \cos( \frac{x}{9} + 55) | + C$

$\int\limits \frac{3dx}{25 {x}^{2} + 81 } = \int\limits \frac{3dx}{ {(5x)}^{2} + 81} \\ \\ 5x = t \\ 5dx = dt \\dx = \frac{dt}{5} \\ \\ \frac{1}{5} \int\limits \frac{3dt}{ {t}^{2} + 81} = \frac{3}{5} \int\limits \frac{dt}{t {}^{2} + {9}^{2} } = \\ = \frac{3}{5} \times \frac{1}{9} \arctg( \frac{t}{9} ) + C = \frac{1}{15} \arctg ( \frac{5x}{9}) + C$

$\int\limits \frac{e {}^{3x} dx}{ \sqrt{5 + e {}^{6x} } } = \int\limits \frac{e {}^{3x} dx}{ \sqrt{5 + {(e {}^{3x} )}^{2} } } \\ \\ e {}^{3x} = t \\ e {}^{3x} \times 3dx = dt \\ {e}^{3x} dx = \frac{dt}{3} \\ \\ \frac{1}{3} \int\limits \frac{dt}{ \sqrt{5 + t {}^{2} } } = \frac{1}{3} \int\limits \frac{dt}{ \sqrt{ {t}^{2} + {( \sqrt{5}) }^{2} } } = \\ = \frac{1}{3} ln |t + \sqrt{5 + {t}^{2} } | + C = \\ = \frac{1}{3} ln | e {}^{3x} + \sqrt{5 + e {}^{6x} } | + C$

$\int\limits \frac{ {x}^{7}dx }{ \sin {}^{2} ( {x}^{8} + 5 ) } \\ \\ {x}^{8} + 5 = t \\ 8 {x}^{7} dx=dt \\ {x}^{7} dx= \frac{dt}{8} \\ \\ \frac{1}{8} \int\limits \frac{dt}{ \sin {}^{2} (t) } = - \frac{1} {8} ctg(t) + C = \\ = - \frac{1}{8} ctg( {x}^{8} + 5) + C$

$\int\limits \frac{9dx}{ {x}^{2} - 8x + 33 } \\ \\ {x}^{2} - 8x + 33 = {x}^{2} - 2 \times x \times 4 + 16 + 17 = \\ = {(x - 4)}^{2} + 17 = {(x - 4)}^{2} + {( \sqrt{17} )}^{2} \\ \\ \int\limits \frac{9dx}{ {(x - 4)}^{2} + {( \sqrt{17}) }^{2} } \\ \\ x - 4 = t \\ dx = dt \\ \\ \int\limits \frac{9dt}{t {}^{2} + {( \sqrt{17} )}^{2} } = \frac{9}{ \sqrt{17} } \arctg( \frac{t}{ \sqrt{17} )} + C = \\ = \frac{9}{ \sqrt{17} } \arctg( \frac{x - 4}{ \sqrt{17} } ) + C$

$\int\limits(1 - 2x) {7}^{9x} dx \\$

По частям:

$U= 1 - 2x \: \: \: \: \: \: dU = (1 - 2x)' dx= - 2dx \\ dV = {7}^{9x} dx \: \: \: \: V = \frac{1}{9} \int\limits {7}^{9x} d(9x) = \frac{1}{9} \times \frac{ {7}^{9x} }{ ln(7) } \\ \\ UV - \int\limits \: VdU = \\ = \frac{1 - 2x}{9 ln(7) } \times {7}^{9x} + \frac{2}{9 ln(7) } \int\limits {7}^{9x} dx = \\ = \frac{(1 - 2x) {7}^{9x} }{9 ln(7) } + \frac{2}{9 ln(7) } \times \frac{ {7}^{9x} }{9 ln(7) } + C = \\ = \frac{7 {}^{9x} }{9 ln(7) } \times (1 - 2x + \frac{2}{9 ln(7) } ) + C = \\ = \frac{ {7}^{9x} }{9 ln(7) } ( \frac{9 ln(7) + 2}{9 ln(7) } - 2x) + C$

$\int\limits \frac{ ln(7x) }{ {x}^{2} } dx \\$

По частям:

$U = ln(7x) \: \:\: \: dU = \frac{1}{7x} \times 7dx = \frac{dx}{x} \\ dV= \frac{dx}{ {x}^{2} } \: \: \:\:\:V = \frac{ {x}^{ - 1} }{ - 1} = - \frac{1}{x} \\ \\ - \frac{ ln(7x) }{x} + \int\limits \frac{1}{x} \times \frac{dx}{x} = \\ = - \frac{ ln(7x) }{x} + \frac{ {x}^{ - 1} }{( - 1)} + C = \\ = - \frac{ ln(7x) }{x} - \frac{1}{x} +C = - \frac{1}{x} ( ln(7x) + 1) + C$