Question

Вычислите определенный интеграл: А) $\int\limits^2_1 \frac{1-x^{6} }{x^{5} } dx$

Б) $\int\limits^2_{-1} \frac{xdx}{(3+x^{2}) ^{3} }$

Answer 1

a

$\int\limits^{ 2 } _ {1} \frac{1 - {x}^{6} }{ {x}^{5} } dx = \int\limits^{ 2 } _ {1}( \frac{1}{ {x}^{5} } - \frac{ {x}^{6} }{ {x}^{5} } )dx \\ = \int\limits^{ 2 } _ {1}( {x}^{ - 5} - x)dx = ( \frac{ {x}^{ - 4} }{( - 4)} - \frac{ {x}^{2} }{2} ) | ^{2} _ {1} = \\ = ( - \frac{1}{4 {x}^{4} } - \frac{ {x}^{2} }{2} )| ^{2} _ {1} = \\ = - \frac{1}{4 \times 16} - 2 + \frac{1}{4} + \frac{1}{2} = \\ = \frac{3}{4} - \frac{1}{64} - 2 = \frac{48 - 1}{64} - 2 = \\ = \frac{47 - 64}{64} = - \frac{17}{64}$

б

$\int\limits^{ 2 } _ { - 1} \frac{xdx}{ {(3 + {x}^{2} )}^{3} } = \frac{1}{2} \int\limits^{ 2} _ { - 1} \frac{2xdx}{ {( {x}^{2} + 3)}^{3} } = \\ =\int\limits^{ 2 } _ { - 1} {(3 + {x}^{2}) }^{ - 3} d(3 + {x}^{2}) = \\ = \frac{ {(3 + {x}^{2}) }^{ - 2} }{ - 2} | ^{2} _ { - 1} = - \frac{1}{2 {(3 + {x}^{2}) }^{2} } | ^{2} _ { - 1} = \\ = - \frac{1}{2 {(3 + 4)}^{2} } + \frac{1}{2 {(3 + 1)}^{2} } = \\ = - \frac{1}{2 \times 49} + \frac{1}{2 \times 16} = \frac{ - 16 + 49}{2 \times 49 \times 16} = \frac{33}{1568}$