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Evaluate the integral ((1/v^9)+(10/v))dv
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anti derivative of \(\frac{1}{v}\) is \(\ln(v)\) as a start
\[\int\limits (\frac{1}{v^9}+\frac{10}{v})dv\] \[\int\limits (v^{-9}+\frac{10}{v})dv\] \[\large =-\frac{1}{8}v^{-8}+10\ln v +C\]
\[\large =-\frac{1}{8v^8}+10\ln v +C\]
@Azteck can you help me with (-8x+8)^-2dx I'm getting (-8x+8)^-1/8 +c ( i don't think this is wrong
@Azteck right**
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You left out a minus sign.
\[-\frac{1}{8}(-8x+8)^{-1}+C\] \[-\frac{1}{8(-8x+8)}+C\]
@Azteck Thanks
No worries.
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