To solve the integral $\int x\sqrt{x-8} \, dx$, we use substitution. Here are the steps:

Step 1: Substitution

Let: $u = x - 8 \quad \text{so that} \quad x = u + 8 \quad \text{and} \quad dx = du.$

Now substitute these into the integral: $\int x\sqrt{x-8} \, dx = \int (u + 8)\sqrt{u} \, du.$

Step 2: Expand the Expression

Split the integral into two parts: $\int (u + 8)\sqrt{u} \, du = \int u\sqrt{u} \, du + \int 8\sqrt{u} \, du.$

Rewrite each term:

• $u\sqrt{u} = u^{3/2}$,
• $8\sqrt{u} = 8u^{1/2}$.

Thus, the integral becomes: $\int u^{3/2} \, du + \int 8u^{1/2} \, du.$

Step 3: Integrate Each Term

For $\int u^{3/2} \, du$: $\int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5}u^{5/2}.$

For $\int 8u^{1/2} \, du$: $\int 8u^{1/2} \, du = 8 \cdot \frac{u^{3/2}}{3/2} = \frac{16}{3}u^{3/2}.$

Step 4: Combine Results

Combine the results: $\int x\sqrt{x-8} \, dx = \frac{2}{5}u^{5/2} + \frac{16}{3}u^{3/2} + C,$ where $C$ is the constant of integration.

Step 5: Substitute Back $u = x - 8$

Replace $u$ with $x - 8$: $\int x\sqrt{x-8} \, dx = \frac{2}{5}(x-8)^{5/2} + \frac{16}{3}(x-8)^{3/2} + C.$

Final Answer:

$\boxed{\int x\sqrt{x-8} \, dx = \frac{2}{5}(x-8)^{5/2} + \frac{16}{3}(x-8)^{3/2} + C.}$

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Related Questions:

1. What are common substitution techniques for integrals involving square roots?
2. How do you handle integrals with powers and roots simultaneously?
3. Can this integral be evaluated using numerical methods instead of symbolic integration?
4. What are some real-world applications of this type of integral?
5. How can we verify the result of this integral by differentiation?

Tip:

When dealing with square roots in integrals, substitution often simplifies the process significantly by eliminating the root.