Question

Найти интегралы с помощью замены переменной

Answer 1

Ответ:

1.

$\int\limits \: x {(x - 1)}^{8} dx$

$x - 1 = t \\ dx = dt \\ x = t + 1$

$\int\limits(t + 1) {t}^{8} dt = \int\limits( {t}^{9} + {t}^{8}) dt = \\ = \frac{ {t}^{10} }{10} + \frac{ {t}^{9} }{9} + c = \\ = \frac{ {(x - 1)}^{10} }{10} + \frac{ {(x - 1)}^{9} }{9} + c$

2.

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

$x + 2 = t \\ dx = dt \\ x = t - 2$

$\int\limits \frac{ {(t - 2)}^{3} }{ {t}^{3} } dt = \int\limits \frac{( {t}^{2} - 4t + 4)(t - 2) }{ {t}^{3} } dt = \\ = \int\limits \frac{ {t}^{3} - 2 {t}^{2} - 4 {t}^{2} + 8t + 4t - 8 }{ {t}^{3} } dt = \\ = \int\limits \frac{ {t}^{3} - 6 {t}^{2} + 12t - 8 }{ {t}^{3} } dt = \\ = \int\limits(1 - \frac{6}{t} + 12 {t}^{ - 2} - 8 {t}^{ -3 } )dt = \\ = t - 6 ln(t) + 12 \times \frac{ {t}^{ - 1} }{( - 1)} - 8 \times \frac{ {t}^{ - 2} }{( - 2)} + c = \\ = t - 6 ln(t) - \frac{12}{t} + \frac{4}{ {t}^{2} } + c = \\ = x + 2 - 6 ln(x + 2) - \frac{12}{x + 2} + \frac{4}{ {(x + 2)}^{2} } + c$

3.

$\int\limits \frac{ {e}^{2x} }{ \sqrt[4]{ {e}^{x} + 1} }$

${e}^{x} + 1 = {t}^{4} \\ {e}^{x} = {t}^{4} - 1 \\ {e}^{2x} = {( {t}^{4} - 1)}^{2} \\ {e}^{x} dx = 4 {t}^{3} dt$

$\int\limits \frac{ {( {t}^{4} - 1)}^{2} \times 4 {t}^{3}dt }{t} = \\ =4 \int\limits {t}^{2} {(t}^{4} - 1)^{2} dt = 4\int\limits {t}^{2} ( {t}^{8} - 2 {t}^{4} + 1)dt = \\ = 4\int\limits( {t}^{10} - 2 {t}^{6} + {t}^{2} )dt = \\ = 4 \times \frac{ {t}^{11} }{11} - 8 \times \frac{ {t}^{7} }{7} + 4 \times \frac{ {t}^{3} }{3} + c = \\ = \frac{4}{11} \sqrt[4]{ {( {e}^{x} + 1)}^{11} } - \frac{8}{7} \sqrt[4]{ {( {e}^{x} + 1)}^{7} } + \frac{4}{3} \sqrt[4]{ {( {e}^{x} + 1) }^{3} } + c = \\ = \sqrt[4]{ {( {e}^{x} + 1) }^{3} } ( \frac{4}{11} {( {e}^{x} + 1)}^{2} - \frac{8}{7} ( {e}^{x} + 1) + \frac{4}{3} ) + c$