Find the indefinite integral.
$\int \frac{x}{\sqrt{x - 9}} d x$

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Final Answer: $\frac{2}{3} (x - 9)^{\frac{3}{2}} + C$

Steps

Step 1 ：Given the integral $\int \frac{x}{\sqrt{x - 9}} d x$

Step 2 ：Let's use substitution to simplify it. Let $u = x - 9$ , then $d u = d x$ . The integral becomes $\int u^{\frac{1}{2}} d u$

Step 3 ：This is a standard form and can be integrated directly. The integral of $u^{\frac{1}{2}}$ with respect to $u$ is $\frac{2}{3} u^{\frac{3}{2}}$

Step 4 ：However, we need to convert it back to the original variable $x$ . So the final answer should be $\frac{2}{3} (x - 9)^{\frac{3}{2}} + C$ , where $C$ is the constant of integration

Step 5 ：Final Answer: $\frac{2}{3} (x - 9)^{\frac{3}{2}} + C$

Find the indefinite integral.∫xx−9dx

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Find the indefinite integral.
$\int \frac{x}{\sqrt{x - 9}} d x$