Question

Calcule $\int\limits {\frac{x^3+1}{x^3-x^2-2x} } \, dx$

Answer 1

Explicação passo-a-passo:

Efectuemos primeiro a divisão dos polinomios

$\frac{ {x}^{3} + 1 }{ {x}^{3} - {x}^{2} - 2x } \\ = \frac{(x + 1)( {x}^{2} - x + 1)}{x( {x}^{2} - x - 2)} \\ = \frac{(x + 1)( {x}^{2} - x + 1) }{x(x + 1)(x - 2)} \\ = \frac{ {x}^{2} - x + 1 }{x(x - 2)} \\ = \frac{(x + 1)(x - 2)}{x(x - 2)} + \frac{3}{x(x - 2)} = 1 + \frac{1}{x} + \frac{3}{x(x - 2)}$

$\frac{ {x}^{3} + 1 }{ {x}^{3} - {x}^{2} - 2x } \\ = \frac{(x + 1)( {x}^{2} - x + 1)}{x( {x}^{2} - x - 2)} \\ = \frac{(x + 1)( {x}^{2} - x + 1) }{x(x + 1)(x - 2)} \\ = \frac{ {x}^{2} - x + 1 }{x(x - 2)} \\ = \frac{(x + 1)(x - 2)}{x(x - 2)} + \frac{3}{x(x - 2)} = 1 + \frac{1}{x} + \frac{3}{x(x - 2)}$

$\int \: (1 + \frac{1}{x} + \frac{3}{x(x - 2)} )dx \\ = x + ln(x) + \int( \frac{ - \frac{3}{2} }{x} + \frac{ \frac{3}{2} }{x - 2})dx \\ = x - \frac{1}{2} ln(x) + \frac{3}{2} ln(x - 2) + c$