Question

решить :

Нужно решить неопределенное интегралы
штук 5-7 на выбор​

Answer 1

1.

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

2.

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

3.

$\int\limits \: tg( \frac{x}{9} + 55)dx = 9\int\limits \: tg( \frac{x}{9} + 55)d( \frac{x}{9}) = \\ = 9\int\limits tg( \frac{x}{9} + 55)d( \frac{x}{9} + 55) = \\ = 9\int\limits \frac{ \sin( \frac{x}{9} + 55) }{ \cos( \frac{x}{9} + 55) } d( \frac{x}{9} + 55) = \\ = - 9\int\limits \frac{d( \cos( \frac{x}{9} + 55)) }{ \cos( \frac{x}{9} + 55)) } = - 9 ln( \cos( \frac{x}{9} + 55 ) ) + C$

4.

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

5.

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

6.

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

7.

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

8.

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

Решаем по частям:

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

9.

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

По частям:

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