Question

Calcule as integrais. (fracoes parciais) 1/x²-4.

Answer 1

Resposta:

$\frac{1}{x^2-4}=\frac{1}{)(x-2)(x+2)}=\frac{A}{x-2}+\frac{B}{x+2} =\frac{A(x+2)+B(x-2)}{(x-2)(x+2)} \\\\ A(x+2)+B(x-2)=1$

P/x = -2 ⇒ A(-2+2) + B(-2-2) = 1

-4B = 1 ⇒ B = -1/4

P/ x = 2 ⇒ A(2 + 2) + B(2 - 2) = 1

4A = 1

A = 1/4

$\frac{A}{x-2} +\frac{B}{x+2}=\frac{\frac{1}{4} }{x-2} +\frac{\frac{-1}{4} }{x+2}$

$\displaystyle\int \ \frac{1}{x^2}dx= \displaystyle\int (\frac{\frac{1}{4} }{x-2} +\frac{\frac{-1}{4} }{x+2}) dx =\frac{1}{4}ln(x-2)-\frac{1}{4}ln(x+2)+c$