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#### renyikouniao

##### Member

- Jun 1, 2013

- 41

integral(upper bound:infinity lower bound:2) 1/(1-x^2) dx

Here's my solution for 3),but I think something went wrong

For all x>=2

0<=-(2-2x)<=-(1-x^2) that means: 0<=-1/(1-x^2)<=-1/(2-2x)

and

integral (upper bound : infinity.lower bound : 2)

-1/(2-2x)=1/2lim(t->infinity)[In(2-2t)-In(-2)]

According to Comparison Test

If ) the integral -1/(2-2x) is convergent,then the integral -1/(1-x^2) is convergent.

But the integral -1/(2-2x) is divergent....so I don't know what to do next...