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OpenStudy (anonymous):
Improper integral: Evaluate the integral from infinity to 1 of 1/[x(sq.rt of x^2 - 1)]
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OpenStudy (anonymous):
$1/x( sqrt x^2 -1 ) dx = arc sec x + c
OpenStudy (anonymous):
thanks i didn't see that it was arcsin!
OpenStudy (anonymous):
i mean arcsec haha
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