Solving a deceptively simple $u$-substitution integral

Question

The integral is $$\int \frac{1}{\sqrt{2x}}\ dx$$

The assignment says use $u$-substitution, but I don't know what I should define $u$ as. Any help would be appreciated, thanks!

Ohh, I get it now! Thanks to everyone who answered!

Answer 1

Set $u=2x$.Then the integral is transformed to $\frac {1}{2} \int \frac {du}{\sqrt{u}}$ which is now easy to solve.

Answer 2

Start by factoring out the $\sqrt{2}$, such that your integral is: $$I=\dfrac{1}{\sqrt{2}}\int\dfrac{1}{\sqrt{x}}dx$$ Now make the substitution: $$u=\sqrt{x}$$ $$du=\dfrac{dx}{2\sqrt{x}}$$ Such that: $$I=\dfrac{2}{\sqrt{2}}\int du=\sqrt{2}\ u$$ And finally, substituting $x$ back: $$\int\dfrac{1}{\sqrt{2x}}dx=\sqrt{2x}$$

Answer 3

I would use $2x = u^2$ Then you get $$ \int \frac{1}{\sqrt{u^2}}udu = \int du = u + C $$ Thus $$ \sqrt{2x} + C $$