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$\int{\frac{8y}{4-y^2}dy}$

The answer isn't in the back of my book, so I have no way to see if I'm right! (I'm about 99% sure I'm wrong though)

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    Yes, you DO have a way to see if you're right. Whatever answer you got - differentiate it, and see if you come up with the original integrand.2012-07-26

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$\int\frac{8y}{4-y^2}\,dy=-4\int\frac{d(4-y^2)}{4-y^2}=-4\log|4-y^2|+K\,\,(constant)$

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    And @Kudla69: in these questions you **always** have a way to check whether you got it right or wrong. Simply...differentiate!2012-07-26
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Letting $u = 4- y^2$, we have

$\int \frac{8y}{4-y^2}dy = -4 \int \frac{du}{u} = -4\log|u| + C = -4\log|4-y^2| + C $