Find the indefinite integral. \[ \int \frac{x}{\sqrt{x+3}} d x \] \[ \int \frac{x}{\sqrt{x+3}} d x=\square \]

Step 1 ：Let \(u = x + 3\). Then, \(du = dx\) and \(x = u - 3\).

Step 2 ：Substitute these values into the integral: \[\int \frac{x}{\sqrt{x+3}} dx = \int \frac{u - 3}{\sqrt{u}} du\]

Step 3 ：Split the fraction into two parts: \[\int \frac{u - 3}{\sqrt{u}} du = \int \frac{u}{\sqrt{u}} du - \int \frac{3}{\sqrt{u}} du\]

Step 4 ：Simplify the integrals: \[\int \frac{u}{\sqrt{u}} du - \int \frac{3}{\sqrt{u}} du = \int u^{1/2} du - 3 \int u^{-1/2} du\]

Step 5 ：Now, we can integrate: \[\int u^{1/2} du - 3 \int u^{-1/2} du = \frac{2}{3} u^{3/2} - 6u^{1/2} + C\]

Step 6 ：Substitute \(u = x + 3\) back into the equation: \[\frac{2}{3} u^{3/2} - 6u^{1/2} + C = \frac{2}{3} (x + 3)^{3/2} - 6(x + 3)^{1/2} + C\]

Step 7 ：\(\boxed{\frac{2}{3} (x + 3)^{3/2} - 6(x + 3)^{1/2} + C}\) is the indefinite integral of \(\int \frac{x}{\sqrt{x+3}} dx\)