Calculate the integral `int_(-5)^(5) (x^(5) sin^(2) x)/( x^(4) + 2x^(2) + 1) dx`

Answer

To solve the integral

\[ I = \int_{-5}^{5} \frac{x^5 \sin^2 x}{x^4 + 2x^2 + 1} \, dx, \]

we can use the property of definite integrals that states:

\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx. \]

### Step 1: Calculate \( f(-x) \)

First, we need to compute \( f(-x) \):

\[ f(x) = \frac{x^5 \sin^2 x}{x^4 + 2x^2 + 1}. \]

Now substituting \(-x\) into \(f(x)\):

\[ f(-x) = \frac{(-x)^5 \sin^2(-x)}{(-x)^4 + 2(-x)^2 + 1}. \]

### Step 2: Simplify \( f(-x) \)

Now simplify \(f(-x)\):

- The numerator becomes:
\[ (-x)^5 = -x^5 \quad \text{and} \quad \sin^2(-x) = \sin^2 x. \]
Thus, the numerator is:
\[ -x^5 \sin^2 x. \]

- The denominator becomes:
\[ (-x)^4 + 2(-x)^2 + 1 = x^4 + 2x^2 + 1. \]

Putting it all together, we have:

\[ f(-x) = \frac{-x^5 \sin^2 x}{x^4 + 2x^2 + 1}. \]

### Step 3: Combine \( f(x) \) and \( f(-x) \)

Now, we can find \( f(x) + f(-x) \):

\[ f(x) + f(-x) = \frac{x^5 \sin^2 x}{x^4 + 2x^2 + 1} + \frac{-x^5 \sin^2 x}{x^4 + 2x^2 + 1}. \]

This simplifies to:

\[ f(x) + f(-x) = \frac{x^5 \sin^2 x - x^5 \sin^2 x}{x^4 + 2x^2 + 1} = 0. \]

### Step 4: Evaluate the integral

Since \( f(x) + f(-x) = 0 \), we have:

\[ I = \int_{-5}^{5} f(x) \, dx = \int_{0}^{5} [f(x) + f(-x)] \, dx = \int_{0}^{5} 0 \, dx = 0. \]

Thus, the value of the integral is:

\[ \boxed{0}. \]