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anti derivative of (x^-6+1/9sqrtx)
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\[\int\limits x^{-6}+{1\over9}\sqrt xdx\]right?
\[(x ^{-6}+\left(\begin{matrix}1 \\ \sqrt[9]{x} \end{matrix}\right)\]
convert the radical expression to a fractional exponent and the problem is very straightforward.
how do i do that?
Did you forget how to write \[{1 \over \sqrt[9]x}\]as a fractional exponent? Remember \[x^{-a}={1\over x^a}\] and \[x^{a/b}=\sqrt[b]{x^a}\]That should help you out here.
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soo x^1/9?? sorry my brain is fried..
\[x^{-1/9} \]because it is in the denominator
to find the der tho i need to bring it up correct?
der?
ya i have to find the anti der of the problem
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anti derivative=integral, same thing you don't HAVE to bring it up, but it makes the answer easier to see.
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