integral 3 to 0 x^3 square root of (x^2+9) dx

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} -

Explore how to solve the integral of x^3 sqrt(x^2 + 9) from 0 to 3 using the substitution method. This detailed solution covers the integration technique by substitution, suitable for advanced college-level students.

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