How do you evaluate $int (x^2-1)/( x+1) dx$ for [1, 2]?

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Answer 1 · 2015-09-19T07:11:25

$1/2$

$int_1^2(x^2-1)/(x+1)dx=int_1^2((x+1)(x-1))/(x+1)dx$
= $int_1^2(x-1)dx=x^2/2-x|_1^2$
= $2^2/2-2-(1/2-1)$
= $1/2$

How do you evaluate int (x^2-1)/( x+1) dxint (x^2-1)/( x+1) dx for [1, 2]?