Use a change of variables to evaluate the following indefinite integral. \( \int \frac{(\sqrt{x}+6)^{8}}{2 \sqrt{x}} d x \) Determine a change of variables from \( x \) to u. Choose the correct answer below. A. \( u=(\sqrt{x}+6)^{8} \) B. \( u=\frac{1}{2 \sqrt{x}} \) C. \( u=\sqrt{x} \) D. \( u=\sqrt{x}+6 \) Write the integral in terms of \( u \). \( \int \frac{(\sqrt{x}+6)^{8}}{2 \sqrt{x}} d x=\int\left(u^{8}\right) \) du Evaluate the integral. \( \int \frac{(\sqrt{x}+6)^{8}}{2 \sqrt{x}} d x=\square \)

Let \( u=\sqrt{x}+6 \). Notice that \[ \frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}, \] so differentiating \( u \) with respect to \( x \) we have \[ \frac{du}{dx}=\frac{1}{2\sqrt{x}}, \] which implies \[ du=\frac{dx}{2\sqrt{x}} \quad \Longrightarrow \quad dx=2\sqrt{x}\,du. \] Substituting into the original integral \[ \int \frac{(\sqrt{x}+6)^8}{2\sqrt{x}}\,dx, \] we replace \((\sqrt{x}+6)^8\) with \(u^8\) and \(dx=2\sqrt{x}\,du\): \[ \int \frac{u^8}{2\sqrt{x}} \times 2\sqrt{x}\,du = \int u^8\,du. \] This shows that option D, \( u=\sqrt{x}+6 \), is the correct substitution. Now, evaluating the integral: \[ \int u^8\,du=\frac{u^9}{9}+C. \] Substituting back \( u=\sqrt{x}+6 \), we obtain \[ \frac{(\sqrt{x}+6)^9}{9}+C. \] Thus, the evaluated integral is \[ \int \frac{(\sqrt{x}+6)^8}{2\sqrt{x}}\,dx=\frac{(\sqrt{x}+6)^9}{9}+C. \]