Evaluate the following integral. \( \int \frac{x}{\sqrt{x-3}} d x \) Click to view the table of general integration formulas. \( \int \frac{x}{\sqrt{x-3}} d x=\square \)

To evaluate the integral \( \int \frac{x}{\sqrt{x-3}} \, dx \), we can use the substitution method. Let \( u = \sqrt{x - 3} \). Therefore, \( x = u^2 + 3 \) and \( dx = 2u \, du \). Now, substituting in: \[ \int \frac{x}{\sqrt{x-3}} \, dx = \int \frac{u^2 + 3}{u} \cdot 2u \, du = 2 \int (u^2 + 3) \, du = 2 \left( \frac{u^3}{3} + 3u \right) + C = \frac{2u^3}{3} + 6u + C \] Substituting back for \( u \): \[ = \frac{2}{3}(x-3)^{3/2} + 6\sqrt{x-3} + C \] So, the final answer is: \[ \int \frac{x}{\sqrt{x-3}} \, dx = \frac{2}{3}(x-3)^{3/2} + 6\sqrt{x-3} + C \]