Evaluate : `int_(0)^(1) cot^(-1) (1-x + x^(2))dx`

Answer

To evaluate the integral \( I = \int_{0}^{1} \cot^{-1}(1 - x + x^2) \, dx \), we can follow these steps:

### Step 1: Use the identity for cotangent inverse
We can use the identity that relates cotangent inverse to tangent inverse:
\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \]
Thus, we can rewrite the integral as:
\[ I = \int_{0}^{1} \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx \]

### Step 2: Simplify the argument of tangent inverse
Next, we simplify the expression \( 1 - x + x^2 \):
\[ 1 - x + x^2 = x^2 - x + 1 \]
This can be factored or rewritten, but it’s already in a suitable form for integration.

### Step 3: Use the identity for the sum of angles
We can use the identity for the sum of angles in tangent inverse:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
We will express \( \tan^{-1}(x) + \tan^{-1}(1 - x) \) in this form.

### Step 4: Set up the integral
We can express the integral as:
\[ I = \int_{0}^{1} \left( \tan^{-1}(x) + \tan^{-1}(1 - x) \right) \, dx \]
Using the property of definite integrals:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(b - x) \, dx \]
This allows us to write:
\[ I = \int_{0}^{1} \tan^{-1}(x) \, dx + \int_{0}^{1} \tan^{-1}(1 - x) \, dx \]

### Step 5: Evaluate the integral
We can evaluate \( \int_{0}^{1} \tan^{-1}(x) \, dx \) using integration by parts:
Let \( u = \tan^{-1}(x) \) and \( dv = dx \).
Then \( du = \frac{1}{1 + x^2} \, dx \) and \( v = x \).

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int_{0}^{1} \tan^{-1}(x) \, dx = \left[ x \tan^{-1}(x) \right]_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x^2} \, dx \]

### Step 6: Evaluate the boundary terms
Evaluating the boundary terms:
\[ \left[ x \tan^{-1}(x) \right]_{0}^{1} = 1 \cdot \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Step 7: Evaluate the remaining integral
Now we need to evaluate:
\[ \int_{0}^{1} \frac{x}{1 + x^2} \, dx \]
Using the substitution \( t = 1 + x^2 \), we find:
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]
The limits change from \( x = 0 \) to \( x = 1 \) which corresponds to \( t = 1 \) to \( t = 2 \):
\[ \int_{1}^{2} \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \ln(t) \bigg|_{1}^{2} = \frac{1}{2} (\ln(2) - \ln(1)) = \frac{1}{2} \ln(2) \]

### Step 8: Combine results
Putting it all together:
\[ \int_{0}^{1} \tan^{-1}(x) \, dx = \frac{\pi}{4} - \frac{1}{2} \ln(2) \]
Thus,
\[ I = 2 \left( \frac{\pi}{4} - \frac{1}{2} \ln(2) \right) = \frac{\pi}{2} - \ln(2) \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{1} \cot^{-1}(1 - x + x^2) \, dx = \frac{\pi}{2} - \ln(2) \]