The value of `int_(0)^(1)(tan^(-1)x)/(cot^(-1)(1-x+x^(2))dx` is____.

Explanation

To solve the integral

\[ I = \int_{0}^{1} \frac{\tan^{-1} x}{\cot^{-1}(1 - x + x^2)} \, dx, \]

we will follow these steps:

### Step 1: Simplify the denominator
We know that

\[ \cot^{-1} y = \frac{\pi}{2} - \tan^{-1} y. \]

Thus, we can rewrite the denominator:

\[ \cot^{-1}(1 - x + x^2) = \frac{\pi}{2} - \tan^{-1}(1 - x + x^2). \]

### Step 2: Rewrite the integral
Substituting this into the integral gives:

\[ I = \int_{0}^{1} \frac{\tan^{-1} x}{\frac{\pi}{2} - \tan^{-1}(1 - x + x^2)} \, dx. \]

### Step 3: Use the property of definite integrals
We can use the property of definite integrals:

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

In our case, \(a = 0\) and \(b = 1\), so we have:

\[ I = \int_{0}^{1} \frac{\tan^{-1}(1 - x)}{\cot^{-1}(1 - (1 - x) + (1 - x)^2)} \, dx. \]

### Step 4: Simplify the new integral
Now, simplify the new integral:

\[ 1 - (1 - x) + (1 - x)^2 = 1 - 1 + x + (1 - 2x + x^2) = x^2 - x + 1. \]

Thus, we have:

\[ I = \int_{0}^{1} \frac{\tan^{-1}(1 - x)}{\cot^{-1}(x^2 - x + 1)} \, dx. \]

### Step 5: Combine the two integrals
Now we have two expressions for \(I\):

1. \( I = \int_{0}^{1} \frac{\tan^{-1} x}{\cot^{-1}(1 - x + x^2)} \, dx \)
2. \( I = \int_{0}^{1} \frac{\tan^{-1}(1 - x)}{\cot^{-1}(x^2 - x + 1)} \, dx \)

Adding these two equations gives:

\[ 2I = \int_{0}^{1} \left( \frac{\tan^{-1} x + \tan^{-1}(1 - x)}{\cot^{-1}(1 - x + x^2)} \right) \, dx. \]

### Step 6: Use the identity for arctangent
Using the identity:

\[ \tan^{-1} x + \tan^{-1}(1 - x) = \frac{\pi}{4}, \]

we can simplify:

\[ 2I = \int_{0}^{1} \frac{\frac{\pi}{4}}{\cot^{-1}(1 - x + x^2)} \, dx. \]

### Step 7: Evaluate the integral
Now, since \(\cot^{-1}(1 - x + x^2)\) is a continuous function over the interval \([0, 1]\), we can evaluate:

\[ 2I = \frac{\pi}{4} \int_{0}^{1} \frac{1}{\cot^{-1}(1 - x + x^2)} \, dx. \]

### Step 8: Solve for \(I\)
Finally, we can solve for \(I\):

\[ I = \frac{1}{2} \int_{0}^{1} 1 \, dx = \frac{1}{2} \cdot (1 - 0) = \frac{1}{2}. \]

Thus, the value of the integral is

\[ \boxed{\frac{1}{2}}. \]