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