Question

Calcule.$\int\limits {\frac{x^5+3}{x^3-4x} } \, dx$

Answer 1

x⁵+3. |x³-4x__

-x⁵+4x³ x²+4

4x³+3

-4x³+16x

16x+3

$\frac{ {x}^{5} +3 }{ {x}^{3} - 4x } = {x}^{2} + 4 + \frac{16x + 3}{ {x}^{3} - 4x }$

$∫\frac{ {x}^{5} +3 }{ {x}^{3} - 4x }dx = \\ ∫ ({x}^{2} + 4)dx + ∫ \frac{16x + 3}{ {x}^{3} - 4x } dx$

$∫ \frac{16x + 3}{x(x - 2)(x + 2)}dx$

$= - \frac{3}{4} ∫ \frac{dx}{x} + \frac{35}{8}∫ \frac{dx}{x - 2} \\ - \frac{29}{8} ∫ \frac{dx}{x + 2}$

$- \frac{3}{4}ln|x|+ \frac{35}{8} ln |x - 2| \\ - \frac{29}{8} ln |x + 2| + c$

$ln | \frac{ {(x - 2)}^{ \frac{35}{8} } }{ {x}^{ \frac{3}{4} } . {(x + 2)}^{ \frac{29}{8} } } | + c$

$∫\frac{ {x}^{5} +3 }{ {x}^{3} - 4x }dx = \\ \frac{ {x}^{3} }{3} + 4x + ln | \frac{ {(x - 2)}^{ \frac{35}{8} } }{ {x}^{ \frac{3}{4} } . {(x + 2)}^{ \frac{29}{8} } } |$

Não deu espaço para colocar a constante mas essa é a resposta.

$\frac{16x + 3}{ {x}^{3} - 4x} = \frac{16x + 3}{x( {x}^{2} - 4) } \\ \frac{16x + 3}{x( {x}^{2} - 4) } = \frac{16x + 3}{x(x - 2)(x + 2)}$

$\frac{16x + 3}{x(x - 2)(x + 2)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}$

$A = \frac{16.0 + 3}{(0 - 2)(0 + 2)} = \frac{3}{( - 2).2} = - \frac{3}{4}$

$B = \frac{16.2 + 3}{2(2 + 2)} = \frac{35}{8}$

$C = \frac{16.( - 2) + 3}{( - 2)( - 2 -2 )} = - \frac{29}{8}$