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

1 Answer

Nov 14, 2015

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

Explanation:

We know that #x^2 - 1 = (x+1)(x-1)#, by the difference of squares so we can rewrite the integral to be

#int((x+1)(x-1))/(x+1)dx = int(x-1)dx = intxdx - intdx = x^2/2 - x + c#

Or if you prefer

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

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