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. $$ \int \frac{(\sqrt{x}+6)^{8}}{2 \sqrt{x}} d x=\square $$

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UpStudy AI · Accepted Answer

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

$$
\int \frac{(\sqrt{x}+6)^8}{2\sqrt{x}}\,dx=\frac{(\sqrt{x}+6)^9}{9}+C.
$$
Let $$ u = \sqrt{x} + 6 $$. Then, the integral becomes $$ \int u^8 \, du $$, which evaluates to $$ \frac{u^9}{9} + C $$. Substituting back, the answer is $$ \frac{(\sqrt{x}+6)^9}{9} + C $$.

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