`int_- 1^1cot^(- 1)((x+x^3)/(1+x^4))dx`

Answer

To solve the integral \( I = \int_{-1}^{1} \cot^{-1}\left(\frac{x+x^3}{1+x^4}\right) dx \), we can use the property of definite integrals. Here’s a step-by-step solution:

### Step 1: Use the property of definite integrals
We know that:
\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(-x) \, dx \]
Thus, we can rewrite our integral as:
\[ I = \int_{0}^{1} \cot^{-1}\left(\frac{x+x^3}{1+x^4}\right) dx + \int_{0}^{1} \cot^{-1}\left(\frac{-x-x^3}{1+x^4}\right) dx \]

### Step 2: Simplify \( \cot^{-1}\left(\frac{-x-x^3}{1+x^4}\right) \)
Using the property of the cotangent inverse function:
\[ \cot^{-1}(-x) = \pi - \cot^{-1}(x) \]
We can rewrite the second integral:
\[ \int_{0}^{1} \cot^{-1}\left(\frac{-x-x^3}{1+x^4}\right) dx = \int_{0}^{1} \left(\pi - \cot^{-1}\left(\frac{x+x^3}{1+x^4}\right)\right) dx \]

### Step 3: Combine the integrals
Now we can combine the two integrals:
\[ I = \int_{0}^{1} \cot^{-1}\left(\frac{x+x^3}{1+x^4}\right) dx + \int_{0}^{1} \left(\pi - \cot^{-1}\left(\frac{x+x^3}{1+x^4}\right)\right) dx \]
This simplifies to:
\[ I = \int_{0}^{1} \cot^{-1}\left(\frac{x+x^3}{1+x^4}\right) dx + \int_{0}^{1} \pi \, dx - \int_{0}^{1} \cot^{-1}\left(\frac{x+x^3}{1+x^4}\right) dx \]
The two \( \cot^{-1} \) integrals cancel out:
\[ I = \int_{0}^{1} \pi \, dx \]

### Step 4: Evaluate the integral
Now we can evaluate the remaining integral:
\[ I = \pi \int_{0}^{1} 1 \, dx = \pi [x]_{0}^{1} = \pi (1 - 0) = \pi \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\pi} \]