Use integral tables to evaluate the integral. \( \int \frac{x}{\sqrt{169+x}} d x \)

To evaluate the integral \( \int \frac{x}{\sqrt{169+x}} \, dx \), we can use the method of substitution and look up relevant integrals. First, let’s make the substitution: \[ u = 169 + x \quad \Rightarrow \quad du = dx \quad \text{and} \quad x = u - 169. \] Then, the integral becomes: \[ \int \frac{u - 169}{\sqrt{u}} \, du = \int \frac{u}{\sqrt{u}} \, du - 169 \int \frac{1}{\sqrt{u}} \, du. \] This simplifies to: \[ \int u^{1/2} \, du - 169 \int u^{-1/2} \, du. \] Now, integrating each term: \[ \int u^{1/2} \, du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3} u^{3/2}, \] \[ \int u^{-1/2} \, du = 2u^{1/2}. \] Thus, we have: \[ \frac{2}{3} u^{3/2} - 169 \cdot 2 u^{1/2} = \frac{2}{3} (169 + x)^{3/2} - 338 (169 + x)^{1/2}. \] Finally, substituting back \( u = 169 + x \): \[ \int \frac{x}{\sqrt{169+x}} \, dx = \frac{2}{3} (169+x)^{3/2} - 338 (169+x)^{1/2} + C, \] where \( C \) is the constant of integration.