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Evaluate `int` 1/(z)^1/25 dz Answer=
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`int 1/z^(1/25) dz`
Now `int x^k dx=1/(k+1) x^(k+1) +c`
So: `int 1/z^(1/25) dz= int z^(-1/25) dz= 1/(-1/25+1) z^(-1/25+1) +c` `=25/24 z^(24/25)+c`
Posted by oldnick on May 8, 2013 at 2:06 AM (Answer #1)
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