QUESTION: a) obtain partial fractions for (x)/((x^2)-1)
b) Use the result of (a) to find the integral of (x^3)/((x^2)-1) dx
(a)
(x)/((x^2)-1) = x/{(x-1)(x+1)}
x/{(x-1)(x+1)} = A/(x-1) + B/(x+1)
multiplying both sides by the denominator (x-1)(x+1),
x = A(x+1) + B(x-1)
x = -1: -1 = 0 +B(-2) => B = 1/2
x = 1: 1 = A(2) + 0 => A = 1/2
Therefore, x/(x^2 - 1) = (1/2){1/(x-1) + 1/(x+1)}
(b)
x^3/(x^2-1) = (x^3 - x + x)/(x^2 - 1) = x(x^2-1)/(x^2-1) + x/(x^2-1) = x + (1/2){1/(x-1) + 1/(x+1)}
Therefore, int (x^3/(x^2-1) dx = int x + (1/2){1/(x-1) + 1/(x+1)} dx
int (x^3/(x^2-1) dx = x^2/2 + (1/2){ln(x-1) + ln(x+1)} = x^2/2 + (1/2)ln(x^2-1)
Answer: (1/2){x^2 + ln(x^2-1)}