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#.