Question

Помогите с интегралами, пожалуйста

Answer 1

Ответ:

1.

$\int\limits {(1 - 4x)}^{3} dx = - \frac{1}{4} \int\limits {(1 - 4x)}^{3} d(1 - 4x) = \\ = - \frac{1}{4} \times \frac{ {(1 - 4x)}^{4} }{4} + C= - \frac{ {(1 - 4x)}^{4} }{16} + C$

2.

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

3.

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

4.

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

5.

$\int\limits \frac{2xdx}{ \sqrt{9 - 8 {x}^{2} } } = \int\limits \frac{d( {x}^{2}) }{ \sqrt{9 - 8 {x}^{2} } } = \\ = \frac{1}{8} \int\limits \frac{d(8 {x}^{2}) }{ \sqrt{9 - 8 {x}^{2} } } = - \frac{1}{8} \int\limits {(9 - 8 {x}^{2} )}^{ - \frac{1}{2 } } d(9 - 8 {x}^{2} ) = \\ = - \frac{1}{8} \times \frac{ {(9 - 8 {x}^{2} )}^{ \frac{1}{2} } }{ \frac{1}{2} } + C= - \frac{ \sqrt{9 - 8 {x}^{2} } }{4} + C$

6.

$\int\limits \frac{dx}{4 {x}^{2} + 3 } = \int\limits \frac{dx}{ {(2x)}^{2} + {( \sqrt{3} )}^{2} } = \\ = \frac{1}{2} \int\limits \frac{d(2x)}{ {(2x)}^{2} + {( \sqrt{3}) }^{2} } = \\ = \frac{1}{2 \sqrt{3} } arctg( \frac{2x}{ \sqrt{3} } ) + C$

7.

$\int\limits \frac{dx}{1 - 3x - {x}^{2} }$

выделим квадрат:

$1 - 3x - {x}^{2} = - ( {x}^{2} + 3x - 1) = \\ = - ( {x}^{2} + 2 \times x \times \frac{3}{2} + \frac{9}{4} - \frac{5}{4} ) = \\ = - ( {(x + \frac{3}{2}) }^{2} - \frac{5}{4} ) = \frac{5}{4} - {(x + \frac{3}{2} )}^{2} = \\ = {( \frac{ \sqrt{5} }{2}) }^{2} - (x + \frac{3}{2} ) {}^{2}$

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

8.

$\int\limits \frac{2x}{3 - 4x - {x}^{2} } dx$

в числителе делаем производную знаменателя:

$(3 - 4x - {x}^{2} )' = - 2x - 4$

$- \int\limits \frac{ - 2xdx}{ - {x}^{2} - 4x + 3 } = - \int\limits \frac{( - 2x - 4 + 4)}{ - {x}^{2} - 4x + 3 }dx = \\ = - \int\limits \frac{( - 2x - 4)dx}{ - {x}^{2} - 4x + 3} - \int\limits \frac{4dx}{ - {x}^{2} - 4x + 3 } = \\ = - \int\limits \frac{d( - {x}^{2} - x4 + 3)}{ - {x}^{2} - 4x + 3} + 4\int\limits \frac{dx}{ {x}^{2} + 4x - 3 } = \\ = - ln(3 - 4x - {x}^{2} ) + 4\int\limits \frac{dx}{ {x}^{2} + 4x + 4 - 7} = \\ = - ln(3 - 4x - {x}^{2} ) + 4\int\limits \frac{d(x + 2)}{ {(x + 2)}^{2} - {( \sqrt{7} )}^{2} } = \\ = - ln(3 - 4x - {x}^{2} ) + \frac{4}{2 \sqrt{7} } ln( \frac{x + 2 - \sqrt{7} }{x + 2 + \sqrt{7} } ) + C = \\ = - ln(3 - 4x - {x}^{2} ) + \frac{2}{ \sqrt{7} } ln( \frac{x + 2 - \sqrt{7} }{x + 2 + \sqrt{7} } ) + C$

9.

$\int\limits \frac{(1 - x)}{ {x}^{2} + 6x + 11 }dx = - \int\limits \frac{x - 1}{ {x}^{2} + 6x + 11} dx = \\ = - \frac{1}{2} \int\limits \frac{2x - 2}{ {x}^{2} + 6x + 11} = - \frac{1}{2} \int\limits \frac{2x + 6 - 8}{ {x}^{2} + 6x + 11} = \\ = - \frac{1}{2} \int\limits \frac{2x + 6}{ {x}^{2} + 6x + 11} dx + \frac{8}{2} \int\limits \frac{dx}{ {x}^{2} + 6x + 11} = \\ = - \frac{1}{2} \int\limits \frac{d( {x}^{2} + 6x + 11) }{ {x}^{2} + 6x + 11 } + 4\int\limits \frac{dx}{ {x}^{2} + 2 \times x \times 3 + 9 + 2 } = \\ = - \frac{1}{2} ln( {x}^{2} + 6x + 11) + 4\int\limits \frac{d(x + 3)}{ {(x + 3)}^{2} + {( \sqrt{2}) }^{2} } = \\ = - \frac{1}{2} ln( {x}^{2} + 6x + 11) + 2 \sqrt{2} arctg( \frac{x + 3}{ \sqrt{2} } ) + C$

10.

$\int\limits \frac{7x + 8}{x(x - 2)} dx$

с помощью неопределенных коэффициентов разделим на простейшие дроби

$\frac{7x + 8}{x(x - 2)} = \frac{A}{x} + \frac{B}{x - 2} \\ 7x + 8 = A(x - 2) + Bx \\ 7x + 8 = Ax - 2A + Bx \\ \\ 7 = A+ B \\8 = - 2 A \\ \\ A= - 4 \\ B = 7 - A = 11 \\ \\ \int\limits \frac{( - 4)dx}{x} + \int\limits \frac{11dx}{x - 2} = \\ = - 4 ln(x) + 11\int\limits \frac{d(x - 2)}{x - 2} = \\ = - 4 ln(x) + 11 ln(x - 2) + C$