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`

integral 1/z^(1/25) dz

= (25 z^(24/25))/24+C

= (25 z^(24/25))/24+C

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