\( \int \frac { 1 } { \sqrt { 3 x } } d x \)

To evaluate the integral \( \int \frac{1}{\sqrt{3x}} \, dx \), we can start with a simple substitution. Notice that you can factor out the square root as follows: \[ \int \frac{1}{\sqrt{3x}} \, dx = \int \frac{1}{\sqrt{3} \sqrt{x}} \, dx = \frac{1}{\sqrt{3}} \int x^{-1/2} \, dx. \] Now, integrating \( x^{-1/2} \) yields: \[ = \frac{1}{\sqrt{3}} \cdot 2 x^{1/2} + C = \frac{2}{\sqrt{3}} \sqrt{x} + C, \] where \( C \) is the constant of integration. So the final result is: \[ \int \frac{1}{\sqrt{3x}} \, dx = \frac{2}{\sqrt{3}} \sqrt{x} + C. \] Taking a step back, integrals like this are part of a fundamental branch of calculus known as antiderivatives. They help us understand how functions behave over intervals, particularly in applications involving area and volume. It's always good to check your work after solving an integral. One common mistake is forgetting to include the constant \( C \) after integrating. Another tip is to be careful with substitutions and limits if you’re dealing with definite integrals; they can change the bounds of the integral differently than expected, especially with non-linear functions.