To solve the integral $\int_{0}^{3} x^3 \sqrt{x^2 + 9} \, dx$, we'll use the substitution method.

Step 1: Substitution

Let $u = x^2 + 9$.

Then, differentiate $u$ with respect to $x$: $\frac{du}{dx} = 2x \quad \Rightarrow \quad du = 2x \, dx$

Now, rewrite the integral in terms of $u$: $x^3 \sqrt{x^2 + 9} \, dx = x^2 \cdot x \sqrt{x^2 + 9} \, dx = \left(\frac{u - 9}{2}\right) \cdot \frac{1}{2} \sqrt{u} \, du$

So, the integral becomes: $\frac{1}{2} \int \left(\frac{u - 9}{2}\right) \sqrt{u} \, du = \frac{1}{4} \int (u - 9) \sqrt{u} \, du$

Step 2: Simplify and Integrate

Expand the integrand: $\frac{1}{4} \int (u \sqrt{u} - 9 \sqrt{u}) \, du = \frac{1}{4} \left( \int u^{3/2} \, du - 9 \int u^{1/2} \, du \right)$

Now, integrate each term: $\int u^{3/2} \, du = \frac{2}{5} u^{5/2}$ $\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$

So the integral becomes: $\frac{1}{4} \left( \frac{2}{5} u^{5/2} - 9 \cdot \frac{2}{3} u^{3/2} \right) + C = \frac{1}{10} u^{5/2} - \frac{3}{2} u^{3/2} + C$

Step 3: Substitute Back $u = x^2 + 9$ and Evaluate

Substitute $u = x^2 + 9$ back into the equation: $\frac{1}{10} (x^2 + 9)^{5/2} - \frac{3}{2} (x^2 + 9)^{3/2} \Bigg|_{0}^{3}$

Evaluate at the bounds $x = 3$ and $x = 0$:

• At $x = 3$: $\frac{1}{10} (3^2 + 9)^{5/2} - \frac{3}{2} (3^2 + 9)^{3/2} = \frac{1}{10} (18)^{5/2} - \frac{3}{2} (18)^{3/2}$
• At $x = 0$: [ \frac{1}{10} (0^2 + 9)^{5/2} -