How do you integrate #int 1/(x^3 - x)# using partial fractions?

1 Answer
George C.
Jan 11, 2017

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

Explanation:

#x^3-x = x(x-1)(x+1)#

So:

#1/(x^3-x) = 1/(x(x-1)(x+1)) = A/x+B/(x-1)+C/(x+1)#

Using Heaviside's cover up method we find:

#A=1/(((color(blue)(0))-1)((color(blue)(0))+1)) = 1/((-1)(1)) = -1#

#B=1/((color(blue)(1))((color(blue)(1))+1))=1/((1)(2))=1/2#

#C=1/((color(blue)(-1))((color(blue)(-1))-1)) = 1/((-1)(-2))=1/2#

So:

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

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

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