To compute the integral of $\frac{x^3}{\sqrt{x^2 + 9}} \, dx$, we can proceed as follows:

Step 1: Simplify the integral

The integral is: $\int \frac{x^3}{\sqrt{x^2 + 9}} \, dx$

Step 2: Substitution

Let: $u = x^2 + 9 \quad \text{so that} \quad du = 2x \, dx$ From $u = x^2 + 9$, we also know that $x^2 = u - 9$. The integral becomes: $\int \frac{x^3}{\sqrt{x^2 + 9}} \, dx = \int \frac{x^2 \cdot x}{\sqrt{u}} \, dx$ Using $x^2 = u - 9$ and $2x \, dx = du$, rewrite as: $\int \frac{(u - 9) x}{\sqrt{u}} \, dx = \int \frac{(u - 9)}{\sqrt{u}} \cdot \frac{du}{2}$

Step 3: Simplify the expression

Distribute and simplify: $\int \frac{(u - 9)}{\sqrt{u}} \cdot \frac{1}{2} \, du = \frac{1}{2} \int \frac{u}{\sqrt{u}} \, du - \frac{9}{2} \int \frac{1}{\sqrt{u}} \, du$ $= \frac{1}{2} \int \sqrt{u} \, du - \frac{9}{2} \int u^{-1/2} \, du$

Step 4: Integrate

For $\int \sqrt{u} \, du$, use the power rule: $\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$ For $\int u^{-1/2} \, du$, use the power rule: $\int u^{-1/2} \, du = 2 u^{1/2}$ Substitute these results: $\frac{1}{2} \cdot \frac{2}{3} u^{3/2} - \frac{9}{2} \cdot 2 u^{1/2}$ $= \frac{1}{3} u^{3/2} - 9 u^{1/2}$

Step 5: Back-substitute $u = x^2 + 9$

Recall that $u = x^2 + 9$. Replace $u$ back: $\frac{1}{3} (x^2 + 9)^{3/2} - 9 (x^2 + 9)^{1/2} + C$

Final Answer:

$\int \frac{x^3}{\sqrt{x^2 + 9}} \, dx = \frac{1}{3} (x^2 + 9)^{3/2} - 9 (x^2 + 9)^{1/2} + C$

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Related Questions:

1. How is substitution used effectively in integration?
2. What is the power rule for integration, and why does it apply here?
3. Can this problem be approached using a trigonometric substitution?
4. What are common mistakes when substituting variables in integrals?
5. How can definite integrals be applied to this expression?

Tip:

Always verify substitutions by differentiating the result to ensure you retrieve the original integrand!