Question

Calcule.$\int\lilimi{\frac{1}{x^2-4} } \, dx$

Answer 1

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

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

$∫ \frac{dx}{ {x}^{2} - 4} = \frac{1}{4}∫ \frac{dx}{x - 2} - \frac{1}{4} ∫ \frac{dx}{x + 2}$

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