# Find the integral: `int_-2^0 (1 / ( 1- 2x + x^2))dx`Answer is shown as: -2/3 . need to know the way how to get there

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`int_-2^0 (1/(1-2x+x^2)) dx`

First we will factor the denominator.

==> `int_-2^0 (1/(x-1)^2) dx`

Let u = x-1 ==> du = dx

`==> int 1/(x-1)^2 dx = int 1/u^2 du = -1/u = -1/(x-1) `

`==> int_-2^0 (1/(x-1)^2) dx = -1/(-2-1) - (-1/(0-1)) = 1/3 -1= -2/3`