## Difficulty to find this integral

1. The problem statement, all variables and given/known data
Hi! I'm not being able to solve this integral: $$\int \sqrt{2ax}\; dx$$ We have started with integral yesterday and I don't know much yet.

2. Relevant equations

3. The attempt at a solution
This is what I tried to solve the integral
$$\int \sqrt{2ax}\; dx = \int u^{\frac{1}{2}}\; du = \frac{2}{3}{}u^{\frac{3}{2}}=\frac{2}{3}\sqrt[3]{(2ax)^2}$$
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 Recognitions: Gold Member Assuming that a is a real constant, $$\int \sqrt{2ax}\; dx = \int \sqrt{2a}\sqrt x\; dx= \sqrt{2a} \int \sqrt x\; dx$$ Since $\sqrt x$ is the same as $x^{\frac{1}{2}}$, you can easily find its integral, and then multiply by $\sqrt {2a}$ to get the final answer.

Mentor
 Quote by DDarthVader 3. The attempt at a solution This is what I tried to solve the integral $$\int \sqrt{2ax}\; dx = \int u^{\frac{1}{2}}\; du = \frac{2}{3}{}u^{\frac{3}{2}}=\frac{2}{3}\sqrt[3]{(2ax)^2}$$
When you use a substitution, as you obviously did above, you need to write down what the substitution is. IOW, you should have u = <...> somewhere close by.

Also, u3/2 = ##\sqrt{u^3}##, not ##\sqrt[3]{u^2}##, which is what you had.