To solve the integral \(\int \frac{1}{\sqrt{2x}} \, dx\), follow these steps:
1. Rewrite the Integral:
[tex]\[
\int \frac{1}{\sqrt{2x}} \, dx
\][/tex]
2. Simplify the Integrand:
Notice that \(\frac{1}{\sqrt{2x}}\) can be rewritten using properties of radicals.
[tex]\[
\frac{1}{\sqrt{2x}} = \frac{1}{\sqrt{2} \cdot \sqrt{x}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{x}}
\][/tex]
3. Factor out the Constant:
Factor out the \(\frac{1}{\sqrt{2}}\), which is a constant, from the integral.
[tex]\[
\int \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{x}} \, dx = \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{x}} \, dx
\][/tex]
4. Recognize the Standard Integral:
The integral \(\int \frac{1}{\sqrt{x}} \, dx\) is a standard integral that equals \(2\sqrt{x}\). Therefore,
[tex]\[
\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{x}} \, dx = \frac{1}{\sqrt{2}} \cdot 2\sqrt{x} = \frac{2\sqrt{x}}{\sqrt{2}}
\][/tex]
5. Simplify the Expression:
Simplify the fraction \(\frac{2\sqrt{x}}{\sqrt{2}}\):
[tex]\[
\frac{2\sqrt{x}}{\sqrt{2}} = \frac{2}{\sqrt{2}} \cdot \sqrt{x} = \sqrt{2} \cdot \sqrt{x} = \sqrt{2x}
\][/tex]
6. Include the Constant of Integration:
Don't forget the constant of integration \(C\), as it is an indefinite integral.
Combining all the steps together, you get:
[tex]\[
\int \frac{1}{\sqrt{2x}} \, dx = \sqrt{2x} + C
\][/tex]
Therefore, the solution to the integral \(\int \frac{1}{\sqrt{2x}} \, dx\) is:
[tex]\[
\sqrt{2x} + C
\][/tex]
