Difficulty to find this integral

[DDarthVader]

Homework Statement

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.

Homework Equations

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}

 

[DryRun]

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.

 

[Mark44]

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