How do you integrate (x^2+2x-1)/(x^3-x)dx?

1 Answer
George C.
Oct 13, 2016

int (x^2+2x-1)/(x^3-x) dx = ln abs(x) + ln abs(x-1) - ln abs(x+1) + C

Explanation:

(x^2+2x-1)/(x^3-x) = (x^2+2x-1)/(x(x-1)(x+1)) = a/x + b/(x-1) + c/(x+1)

Use Heaviside's cover up method to find a, b, c:

a = ((color(blue)(0))^2+2(color(blue)(0))-1)/(((color(blue)(0))-1)((color(blue)(0))+1)) = (-1)/((-1)(1)) = 1

b = ((color(blue)(1))^2+2(color(blue)(1))-1)/((color(blue)(1))((color(blue)(1))+1)) = (2)/((1)(2)) = 1

c = ((color(blue)(-1))^2+2(color(blue)(-1))-1)/((color(blue)(-1))((color(blue)(-1))-1)) = (-2)/((-1)(-2)) = -1

So:

int (x^2+2x-1)/(x^3-x) dx = int 1/x + 1/(x-1) -1/(x+1) dx

color(white)(int (x^2+2x-1)/(x^3-x) dx) = ln abs(x) + ln abs(x-1) - ln abs(x+1) + C

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