To solve the integral \( \int_0^1 \cot^{-1}(1 - x + x^2) \, dx \), we can follow these steps:
### Step 1: Rewrite the Integral
We start by rewriting the cotangent inverse function:
\[
\cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right)
\]
Thus, we can rewrite the integral as:
\[
\int_0^1 \cot^{-1}(1 - x + x^2) \, dx = \int_0^1 \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx
\]
### Step 2: Simplify the Argument
Next, we simplify \( 1 - x + x^2 \):
\[
1 - x + x^2 = x^2 - x + 1
\]
This expression does not factor nicely, but we can proceed with the integral.
### Step 3: Use a Property of Integrals
We can use the property of integrals that states:
\[
\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx
\]
In our case, \( a = 0 \) and \( b = 1 \), so:
\[
\int_0^1 \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx = \int_0^1 \tan^{-1}\left(\frac{1}{1 - (1 - x) + (1 - x)^2}\right) \, dx
\]
This simplifies to:
\[
\int_0^1 \tan^{-1}\left(\frac{1}{x^2 + x}\right) \, dx
\]
### Step 4: Combine the Integrals
Now we can combine the two integrals:
\[
I = \int_0^1 \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx + \int_0^1 \tan^{-1}\left(\frac{1}{x^2 + x}\right) \, dx
\]
Let \( I_1 = \int_0^1 \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx \) and \( I_2 = \int_0^1 \tan^{-1}\left(\frac{1}{x^2 + x}\right) \, dx \).
### Step 5: Evaluate the Combined Integral
Notice that \( I_1 + I_2 \) can be simplified:
\[
I_1 + I_2 = \int_0^1 \left( \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) + \tan^{-1}\left(\frac{1}{x^2 + x}\right) \right) \, dx
\]
Using the identity \( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \) if \( ab < 1 \).
### Step 6: Find the Result
After evaluating the integrals, we find:
\[
I = \frac{\pi}{2}
\]
Thus, the value of the integral is:
\[
\int_0^1 \cot^{-1}(1 - x + x^2) \, dx = \frac{\pi}{2} - \ln(2)
\]
### Final Answer
The value of the integral is:
\[
\frac{\pi}{2} - \ln(2)
\]
