How do you integrate $\int \frac{x^{2} - 1}{x + 1} d x$ $int (x^2-1)/( x+1 )dx$ ?

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How do you integrate $\int \frac{x^{2} - 1}{x + 1} d x$ $int (x^2-1)/( x+1 )dx$ ?

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Answer 1 · 2015-11-14T18:50:12

$\int (\frac{x^{2} - 1}{x + 1}) = \frac{x^{2}}{2} - x + c$ $int((x^2-1)/(x+1)) = (x^2)/2 -x+c$

Explanation:

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

$\int \frac{(x + 1) (x - 1)}{x + 1} d x = \int (x - 1) d x = \int x d x - \int d x = \frac{x^{2}}{2} - x + c$ $int((x+1)(x-1))/(x+1)dx = int(x-1)dx = intxdx - intdx = x^2/2 - x + c$

Or if you prefer

$\int (\frac{x^{2} - 1}{x + 1}) = \frac{x^{2} - 2 x + c}{2}$ $int((x^2-1)/(x+1)) = (x^2-2x+c)/2$

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How do you integrate $\int \frac{x^{2} - 1}{x + 1} d x$ $int (x^2-1)/( x+1 )dx$ ?

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