Найти интегралы, методом непосредственного интегрирования

Ответы

superbest 24.02.2021 15:33

$\int\limits(5 {x}^{3} - 2 {x}^{2} + 3x - 8)dx = \frac{5 {x}^{4} }{4} - \frac{2 {x}^{3} }{3} + \frac{3 {x}^{2} }{2} - 8x + C \\$

$\int\limits(4 {x}^{3} - 15 {x}^{2} + 14x - 3)dx = \frac{4 {x}^{4} }{4} - \frac{15 {x}^{3} }{3} + \frac{14 {x}^{2} }{2} - 3x + C = \\ = {x}^{4} - 5 {x}^{3} + 7 {x}^{2} - 3x + C$

$\int\limits {(2x - 1)}^{3} dx = \frac{1}{2} \int\limits {(2x - 1)}^{3} d(2x) = \\ = \frac{1}{2} \int\limits {(2x - 1)}^{3} d(2x - 1) = \frac{ {(2x - 1)}^{4} }{8} + C$

$\int\limits {x}^{3} (1 + 5x)dx = \int\limits( {x}^{3} + 5 {x}^{4} )dx = \\ = \frac{5 {x}^{5} }{5} + \frac{ {x}^{4} }{4} + c = {x}^{5} + \frac{ {x}^{4} }{4} + C$

$\int\limits \frac{3 {x}^{3} - 2 {x}^{2} + 5x }{2x} dx = \int\limits ( \frac{3}{2} {x}^{2} - x + \frac{5}{2} ) = \\ = \frac{3}{2} \times \frac{ {x}^{3} }{3} - \frac{ {x}^{2} }{2} + 2.5x + C = \\ = \frac{ {x}^{3} }{2} - \frac{ {x}^{2} }{2} + 2.5x + C$

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

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

$\int\limits \frac{ {x}^{3} - 2 {x}^{2} - 3x - 4 }{ {x}^{2} } dx =\int\limits(x - 2 - \frac{3}{x} - \frac{4}{ {x}^{2} } )dx = \\ = \frac{ {x}^{2} }{2} - 2x - 3 ln(x) - 4 \times \frac{ {x}^{ - 1} }{( - 1)} + C = \\ = \frac{ {x}^{2} }{2} - 2x - 3 ln(x) + \frac{4}{x} + C$

$\int\limits \frac{ {(3x + 1)}^{2} }{x} dx =\int\limits \frac{9 {x}^{2} + 6x + 1}{x} dx = \\ = \int\limits(9x + 6 + \frac{1}{x} )dx = \\ = \frac{9 {x}^{2} }{2} + 6x + ln(x) + C$