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Make a substitution to express the integrand as a rational function and then evaluate the integral. Integrate[sqrt x/(x-36), {x, 4, 9}]
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\[\int\limits_{4}^{9} \sqrt x/(x-36) dx\]
|dw:1328291169857:dw|
Substitute \(u=\sqrt{x} \implies x=u^2 \implies dx=2udu \) to get \[\int_4^9 \frac{u}{u^2-36}2udu=2\int_4^9 \frac{u^2}{u^2-36}du=\cdots\] Can you take it from here?
|dw:1328291206918:dw|
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