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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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    You know how to split into partial fractions?2012-07-26
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    hmm, I don't think I do. What do you mean by that?2012-07-26
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    @J.M., no need. Try the substitution $u = 4 - y^2.$ Then $du = -2y \, dy$ and $-4 du = 8 y \, dy$2012-07-26
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    That's actually what I did do, except for your last step I did: $\frac{1}{-2}du = ydy$2012-07-26
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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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    I actually did get it right! :] what did you do in your second step though?2012-07-26
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    I just wrote $\,8y=-4(-2y)\,$ and wrote $\,-2y=d(4-y^2)\,$ , to make crystal clear that we've here an integral of the form $$\int\frac{f'(x)\,dx}{f(x)}=\log|f(x)|+K$$That's all.2012-07-26
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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

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