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 can start by rewriting the cotangent inverse function. Recall that

\[ \cot^{-1} \theta = \tan^{-1} \left( \frac{1}{\theta} \right). \]

Thus, we can express \(\cot^{-1}(1 - x + x^2)\) as:

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

This allows us to rewrite the integral:

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

Next, we can use the property of inverse tangent that states:

\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right), \]

to analyze the denominator. However, for this problem, we will also utilize a symmetry property of definite integrals.

We can define a new integral \( J \):

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

Notice that \(1 - x\) is a reflection about \(x = \frac{1}{2}\). We can rewrite \(J\) as:

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

Now, we can add \(I\) and \(J\):

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

Using the property of \(\tan^{-1}\):

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

we can simplify the integral:

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

Now, we can also observe that:

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

Thus, we can express \(I + J\) as:

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

Since \(I = J\), we have:

\[ 2I = \int_{0}^{1} 1 \, dx = 1. \]

Therefore, we find:

\[ I = \frac{1}{2}. \]

Thus, the value of the integral is:

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