antiderivative of $\frac{4}{\sqrt{u}}$

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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That'd be better if you specify with respect to what eventually you want to integrate. –  Kaster Dec 13 '12 at 2:28
@soniccool: You previous question covers almost exactly the same material. Why don't you spend some time thinking about the answeres given there. –  JavaMan Dec 13 '12 at 2:29

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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Yea this is where im stuck i dont know how to continue now with fractions and the u and what not? –  soniccool Dec 13 '12 at 2:27
I filled in some more details. Let me know if you something does not make sense. –  Joe Dec 13 '12 at 2:31
tHAT FORUMULA HELPED! Awesome now i understandit actually –  soniccool Dec 13 '12 at 2:31
Glad to hear it! Note that the formula only holds for polynomials of the form $x^n$. –  Joe Dec 13 '12 at 2:32

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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I dont get how we get 8 out of that? –  soniccool Dec 13 '12 at 2:28
Look at @Brett Frankel's calculation and see that $4/(1/2)=8.$ –  learner Dec 13 '12 at 2:32
@soniccool Try differentiating the antiderivative and see if that helps you see what happens to the constants out front. –  Brett Frankel Dec 13 '12 at 2:32