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Find the indefinite integral,

$$ I = \int \frac{8 - 2x}{\sqrt{6x - x^2}} dx\,. $$

I know this is a 'substitution' question, but I can't work out what to substitute. Please could you tell me the substitution and how you got to it.

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    $\begingroup$ Complete the square. Inside the square root we have $9-(x-3)^2$. Ultimately we will probably let $x-3=3\sin\theta$. Or else more slowly $x-3=3t$. $\endgroup$ – André Nicolas Jan 11 '15 at 17:59
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    $\begingroup$ It may be handy as a simplifying device to write the top as $6-2x+2$. This is helpful, for $6-2x$ is the derivative of $6x-x^2$. $\endgroup$ – André Nicolas Jan 11 '15 at 18:02
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use the substitution $3 \sin t = x - 3$ which comes from completing the square of $6x - x^2$ and looking ahead so that you can get rid of the square root in the denominator.

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