How do you find the integral of ∫ dx/(1-x)^2 from negative infinity to 0?

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Answer 1 · 2018-01-23T17:39:06

1.

We want to calculate $I = int_-oo^0 dx/(1-x)^2$ .

Let $t = 1-x$ ; then $dt = -dx$ . Now, our new integration limits will be for:

$t = oo$ , when $x = -oo$ ; and

$t = 1$ , when $x = 0$ . Then:

$I = int_oo^1(-dt)/t^2$ ;

$I = int_1^oodt/t^2$ ;

$I = (-1/t)|_1^oo$ ;

$I = [lim_(t->oo)(-1/t) - (-1/1)]$ ;

$I = 1$ .