Question

Найти интегралы , с решением

Answer 1

Пошаговое объяснение:

1

$\int\limits \frac{dx}{ {x}^{2} + 8x + 23} \\ \\ {x}^{2} + 8x + 23 = {x}^{2} + 2 \times x \times 4 + 16 + 7 = \\ = {(x + 4)}^{2} + 7 = {(x + 2)}^{2} + {( \sqrt{7} )}^{2} \\ \\ \int\limits \frac{dx}{ {(x + 2)}^{2} + {( \sqrt{7}) }^{2} } = \int\limits \frac{d(x + 2)}{ {(x + 2)}^{2} + {( \sqrt{7} )}^{2} } = \\ = \frac{1}{ \sqrt{7} } arctg( \frac{x + 2}{ \sqrt{7} } ) + C$

2

$\int\limits \frac{4 - 5x}{5 {x}^{2} - x + 7 } dx \\ \\ \text{Делаем в числителе производную знаменателя:} \\ (5 {x}^{2} - x + 7)' = 10x - 1 \\ \\ - \int\limits \frac{5x - 4}{5 {x}^{2} - x + 7 } dx = - \frac{1}{2} \int\limits \frac{10x - 8}{5 {x}^{2} - x + 7 } dx = \\ = - \frac{1}{2} \int\limits \frac{10x - 1 - 7}{5 {x}^{2} - x + 7} dx = \\ = - \frac{1}{2} \int\limits \frac{10x - 1}{5 {x}^{2} - x + 7} dx+ \frac{7}{2} \int\limits \frac{dx}{5 {x}^{2} - x + 7 } = \\ = - \frac{1}{2} \int\limits \frac{d(5 {x}^{2} - x + 7)}{5 {x}^{2} - x + 7} + \frac{7}{2} \int\limits \frac{dx}{ {( \sqrt{5}x) }^{2} - \sqrt{5} x \times 2 \times \frac{1}{2 \sqrt{5} } + \frac{1}{20} + \frac{139}{20} } = \\ = - \frac{1}{2} ln |5 {x}^{2} - x + 7 | + \frac{7}{2} \times \frac{1}{ \sqrt{5} } \int\limits \frac{( \sqrt{5}x - \frac{1}{2 \sqrt{5} }) }{ {( \sqrt{5} x - \frac{1}{2 \sqrt{5} } )}^{2} + {( \sqrt{ \frac{139}{20} }) }^{2} } = \\ = - \frac{1}{2} ln |5 {x}^{2} - x + 7| + \frac{7}{2 \sqrt{5} } \times \frac{2 \sqrt{5} }{ \sqrt{139} } arctg( \frac{ \sqrt{5}x - \frac{1}{2 \sqrt{5} } }{ \frac{ \sqrt{139} }{2 \sqrt{5} } } ) +C = \\ = - \frac{1}{2} ln |5 {x}^{2} - x + 7| + \frac{7}{ \sqrt{139} } arctg( \frac{10x - 1}{ \sqrt{139} } ) + C$

3

$\int\limits \frac{2x - 3}{ \sqrt{5 + x - 2 {x}^{2} } } dx \\ \\ \text{В числителе делаем производную знаменателя:} \\ (5 + x - 2 {x}^{2} )' = (1 - 4x) \\ \\ - \frac{1}{2} \int\limits \frac{( - 4x + 6)dx}{ \sqrt{5 + x - 2 {x}^{2} } } = - \frac{1}{2} \int\limits \frac{( - 4x + 1 + 5)dx}{ \sqrt{5 + x - 2 {x}^{2} } } = \\ = - \frac{1}{2} \int\limits \frac{(1 -4 x)}{ \sqrt{5 + x - 2 {x}^{2} } } dx - \frac{5}{2} \int\limits \frac{dx}{ \sqrt{5 + x - 2 {x}^{2} } } = \\ = - \frac{1}{2} \int\limits \frac{d(5 + x - 2 {x}^{2} )}{ \sqrt{5 +x - 2 {x}^{2} } } - \frac{5}{2} \int\limits \frac{dx}{ \sqrt{5 + x - 2 {x}^{2} } } \\ \\ 5 + x - 2 {x}^{2} = - (2 {x}^{2} - x - 5) = \\ = - ( {( \sqrt{2} x)}^{2} - \sqrt{2} x \times 2 \times \frac{1}{2 \sqrt{2} } + \frac{1}{8} - \frac{41}{8} ) = \\ = - ( {( \sqrt{2}x - \frac{1}{2 \sqrt{2} } ) }^{2} - ( \frac{ \sqrt{41} }{2 \sqrt{2} } ) {}^{2} ) = \\ = {( \frac{ \sqrt{41} }{2 \sqrt{2} }) }^{2} - {( \sqrt{2} x - \frac{1}{2 \sqrt{2} } )}^{2} \\ \\ = - \frac{1}{2} ln |5 + x - 2 {x}^{2} | - \frac{5}{2} \times \frac{1}{ \sqrt{2} } \int\limits \frac{d( \sqrt{2} x - \frac{1}{2 \sqrt{2} } )}{ {( \frac{ \sqrt{41} }{2 \sqrt{2} } ) {}^{2} - {( \sqrt{2} x - \frac{1}{2 \sqrt{2} } )}^{2} }^{} } = \\ = - \frac{1}{2} ln |5 + x - 2 {x}^{2} | - \frac{5}{2 \sqrt{2} } \times \frac{2 \sqrt{2} }{2 \sqrt{41} } ln | \frac{ \frac{ \sqrt{41} }{2 \sqrt{2} } - \sqrt{2}x + \frac{1}{2 \sqrt{2} } }{ \frac{ \sqrt{41} }{2 \sqrt{2} } + \sqrt{2}x - \frac{1}{2 \sqrt{2} } } | + C= \\ = - \frac{1}{2} ln |5 + x - 2 {x}^{2} | - \frac{5}{2 \sqrt{41} } ln | \frac{ \sqrt{41} + 1 - 8x }{ \sqrt{41} - 1 + 8x} | + C$