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I am trying to get the antiderivative of $\frac{4}{\sqrt{u}}$

Im not sure how to apply antiderivative rules when having a question like this?

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    @soniccool: You [previous question](http://math.stackexchange.com/questions/257535/antiderivative-of-8t-1-2) covers _almost exactly_ the same material. Why don't you spend some time thinking about the answeres given there.2012-12-13

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Hint: Rewrite $\sqrt{u}$ as $u^{1/2}$. Thus, $\frac{4}{\sqrt{u}} = 4u^{-1/2}$.

Recall that $\int x^n \ dx = \frac{x^{n+1}}{n+1}$ for all $n \ne -1$.

Then, $4\int u^{-1/2} \ du = 4 \cdot \frac{u^{-1/2 + 1}}{1/2} = 8u^{1/2} + C = 8\sqrt{u} + C$

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    Glad to hear it! Note that the formula only holds for polynomials of the form $x^n$.2012-12-13
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Hint: You can re-write this as $4 u^{-\frac{1}{2}}$. Now, use the rule for integrating functions of the form $\int u^n du$. I'll give more details if you need.

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The antiderivative of $\frac{4}{\sqrt u}=4u^{-1/2}$ is $4\cdot \frac{1}{1/2}u^{-1/2+1}+C=8u^{1/2}+C$

If you are just looking for an answer (and not an explanation), try this. On the other hand, if you're looking to understand, please make sure you understand each step and comment below if you need further clarification.

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    @soniccool Try differentiating the antiderivative and see if that helps you see what happens to the constants out front.2012-12-13