# Difficulty to find this integral

## 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.

## The Attempt at a Solution

This is what I tried to solve the integral
$$\int \sqrt{2ax}\; dx = \int u^{\frac{1}{2}}\; du = \frac{2}{3}{}u^{\frac{3}{2}}=\frac{2}{3}\sqrt[3]{(2ax)^2}$$

## Answers and Replies

DryRun
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.

Mark44
Mentor

## 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.