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

To evaluate the definite integral 

\[
\int_{0}^{3} x^3 \sqrt{x^2 + 9} \, dx,
\]

we can use the substitution method. Here's a step-by-step solution:

### Step 1: Substitution
Let:

\[
u = x^2 + 9.
\]

Then, the differential of \(u\) with respect to \(x\) is:

\[
\frac{du}{dx} = 2x \implies du = 2x \, dx.
\]

Now, express \(x^3\) and \(dx\) in terms of \(u\):

\[
x^2 = u - 9 \implies x = \sqrt{u - 9}.
\]

Therefore:

\[
x^3 = (x)(x^2) = \sqrt{u - 9} \cdot (u - 9) = (u - 9)^{3/2}.
\]

From the differential:

\[
dx = \frac{du}{2x} = \frac{du}{2\sqrt{u - 9}}.
\]

### Step 2: Change the Limits of Integration
When \(x = 0\):

\[
u = 0^2 + 9 =

Learn how to evaluate the definite integral of x^3 sqrt(x^2 + 9) from 0 to 3 with a detailed step-by-step solution using the substitution method. This problem involves advanced integration techniques and is suitable for advanced high school students.

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