The value of `int_(-1)^1cot^-1((x+x^3+x^5)/(x^4+x^2+1))` dx is equal to

Options

A. `pi/2`
B. `pi/4`
C. `(3pi)/4`
D. `pi`

Answer

To solve the integral

\[ I = \int_{-1}^{1} \cot^{-1}\left(\frac{x+x^3+x^5}{x^4+x^2+1}\right) dx, \]

we can utilize the property of definite integrals that states:

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

### Step 1: Define the function and find \( f(-x) \)

Let

\[ f(x) = \cot^{-1}\left(\frac{x+x^3+x^5}{x^4+x^2+1}\right). \]

Now, we need to find \( f(-x) \):

\[ f(-x) = \cot^{-1}\left(\frac{-x + (-x)^3 + (-x)^5}{(-x)^4 + (-x)^2 + 1}\right) = \cot^{-1}\left(\frac{-x - x^3 - x^5}{x^4 + x^2 + 1}\right). \]

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

Using the identity \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \), we can rewrite \( f(-x) \):

\[ f(-x) = \cot^{-1}\left(-\frac{x+x^3+x^5}{x^4+x^2+1}\right) = \pi - \cot^{-1}\left(\frac{x+x^3+x^5}{x^4+x^2+1}\right) = \pi - f(x). \]

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

Now we can combine \( f(x) \) and \( f(-x) \):

\[ f(x) + f(-x) = f(x) + \left(\pi - f(x)\right) = \pi. \]

### Step 4: Set up the integral

Now we can set up the integral:

\[ I = \int_{-1}^{1} f(x) \, dx = \int_{0}^{1} \left( f(x) + f(-x) \right) \, dx = \int_{0}^{1} \pi \, dx. \]

### Step 5: Evaluate the integral

The integral simplifies to:

\[ I = \int_{0}^{1} \pi \, dx = \pi \cdot x \bigg|_{0}^{1} = \pi \cdot (1 - 0) = \pi. \]

Thus, the value of the integral is

\[ \boxed{\pi}. \]