Question

integral de x/(x2-1)(x-2)

Answer 1

$\displaystyle \int \frac{x}{(x^2-1)(x-2)}dx\\ \\ \\ \text{Fracciones parciales:}\\ \frac{x}{(x^2-1)(x-2)}=\frac{x}{(x-1)(x+1)(x-2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}\\ \\ \\ \frac{x}{(x-1)(x+1)(x-2)} = \frac{x^2(A + B + C) - x(A + 3B) - 2A + 2B - C}{(x-1)(x+1)(x-2)}\\ \\ \\ x=x^2(A + B + C) - x(A + 3B) - 2A + 2B - C$

Compara cada coeficiente y obtienes el siguiente sistema de ecuaciones

$\begin{cases} A + B + C=0\\A + 3B=-1\\- 2A + 2B - C=0 \end{cases}\\ \\\\ A=-\dfrac{1}{2} \\ B=- \dfrac{1}{6}\\ C=\dfrac{2}{3}$

Entonces resta integrar:

$\displaystyle I=\int -\frac{1}{2}\cdot\frac{1}{x-1}-\frac{1}{6}\cdot\frac{1}{x+1}+\frac{2}{3}\cdot\frac{1}{x-2}dx\\ \\ I=-\frac{1}{2} \int \frac{dx}{x-1} -\frac{1}{6} \int \frac{dx}{x+1}+\frac{2}{3}\int\frac{dx}{x-2}\\ \\ \\ \boxed{I=-\frac{1}{2} \ln |x-1|-\frac{1}{6}\ln|x+1|+\frac{2}{3}\ln|x-2|+C}$