Question

integral de (x^3-4x^2+6x-8)/(x*(x^2-2x-8) dx

Answer 1

$\int\frac{x^3-4x^2+6x-8}{x(x^2-2x-8)}dx$

$\frac{x^3-4x^2+6x-8}{x(x^2-2x-8)}=1+\frac{2(x^2 - 7x + 4)}{x(x^2-2x-8)}$

$\frac{x^3-4x^2+6x-8}{x(x^2-2x-8)}=1+\frac{2(x^2 - 7x + 4)}{x(x-4)(x+2)}$

separemos esta fracción propia

$\frac{2(x^2 - 7x + 4)}{x(x-4)(x+2)}$

$\frac{2(x^2 - 7x + 4)}{x(x-4)(x+2)}=\frac{A}{x}+\frac{B}{x-4}+\frac{C}{x+2}$

$x^2(A + B +C) - 2x(A - B + 2C) - 8A = 2x^2-14x+8$

comparemos coeficientes
$A+B+C=2\\ A-B+2C=7\\A=-1$

al resolver se obtiene
$A=-1\\B=-2/3\\C=11/3$

Por ende

$\frac{x^3-4x^2+6x-8}{x(x^2-2x-8)}=1-\frac{1}{x}-\frac{2/3}{x-4}+\frac{11/3}{x+2}$

entonces

$\int \frac{x^3-4x^2+6x-8}{x(x^2-2x-8)}dx=\int1-\frac{1}{x}-\frac{2/3}{x-4}+\frac{11/3}{x+2}dx$

$\int \frac{x^3-4x^2+6x-8}{x(x^2-2x-8)}dx=x-\ln|x|-2/3\ln|x-4|+11/3\ln|x+2|+C$