Use any method to evaluate the integral. $$ \int \frac{x^{3} d x}{\sqrt{x^{2}+100}} $$

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To evaluate the integral:

$$
\int \frac{x^{3} \, dx}{\sqrt{x^{2}+100}},
$$

we can use a substitution method. Let’s proceed step-by-step.

### Step 1: Substitution

Let:
$$
u = x^2 + 100
$$
Then, the derivative of $$ u $$ with respect to $$ x $$ is:
$$
\frac{du}{dx} = 2x \quad \Rightarrow \quad du = 2x \, dx \quad \Rightarrow \quad x \, dx = \frac{du}{2}
$$

### Step 2: Rewrite the Integral in Terms of $$ u $$

Notice that $$ x^3 $$ can be expressed as $$ x^2 \cdot x $$. Since $$ u = x^2 + 100 $$, we have:
$$
x^2 = u - 100
$$
Thus, the integral becomes:
$$
\int \frac{x^{3} \, dx}{\sqrt{x^{2}+100}} = \int \frac{(u - 100) \cdot x \, dx}{\sqrt{u}} = \int \frac{(u - 100)}{\sqrt{u}} \cdot \frac{du}{2}
$$
$$
= \frac{1}{2} \int (u^{1/2} - 100u^{-1/2}) \, du
$$

### Step 3: Integrate Term by Term

Now, integrate each term separately:
$$
\frac{1}{2} \int u^{1/2} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{u^{3/2}}{3}
$$
$$
\frac{1}{2} \int (-100u^{-1/2}) \, du = -50 \cdot 2u^{1/2} = -100u^{1/2}
$$
Combining these results:
$$
\frac{u^{3/2}}{3} - 100u^{1/2} + C
$$
where $$ C $$ is the constant of integration.

### Step 4: Substitute Back $$ u = x^2 + 100 $$

Replace $$ u $$ with $$ x^2 + 100 $$:
$$
\frac{(x^2 + 100)^{3/2}}{3} - 100(x^2 + 100)^{1/2} + C
$$

### Final Answer

$$
\boxed{\,\frac{(x^{2}+100)^{3/2}}{3}\; -\; 100\,\sqrt{x^{2}+100}\;+\;C\,}
$$
The integral evaluates to: $$ \frac{(x^{2}+100)^{3/2}}{3} - 100\sqrt{x^{2}+100} + C $$

Use any method to evaluate the integral.

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