`int_(0)^(1) tan^(-1) (1-x+x^(2)) dx`=

Options

A. log 2
B. log `(1)/(2)`
C. `pi log 2`
D. `(pi)/(2) "log" (1)/(2)`

Answer

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

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{1} \tan^{-1}(1 - x + x^2) \, dx \]

### Step 2: Use the identity for \(\tan^{-1}\)
We can use the identity:
\[ \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2} \]
This allows us to write:
\[ \tan^{-1}(1 - x + x^2) = \frac{\pi}{2} - \cot^{-1}(1 - x + x^2) \]

### Step 3: Rewrite the integral
Substituting this into our integral gives:
\[ I = \int_{0}^{1} \left( \frac{\pi}{2} - \cot^{-1}(1 - x + x^2) \right) \, dx \]
This simplifies to:
\[ I = \frac{\pi}{2} \cdot 1 - \int_{0}^{1} \cot^{-1}(1 - x + x^2) \, dx \]
\[ I = \frac{\pi}{2} - J \]
where \( J = \int_{0}^{1} \cot^{-1}(1 - x + x^2) \, dx \).

### Step 4: Change of variable for \( J \)
Now, we can use the property of integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
Setting \( x = 1 - u \), we have:
\[ J = \int_{0}^{1} \cot^{-1}(1 - (1 - u) + (1 - u)^2) \, du \]
This simplifies to:
\[ J = \int_{0}^{1} \cot^{-1}(u + u^2) \, du \]

### Step 5: Combine \( I \) and \( J \)
Now we have:
\[ I = \frac{\pi}{2} - J \]
and since \( J \) has the same form as \( I \), we can conclude that:
\[ I + J = \frac{\pi}{2} \]
Thus:
\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4} \]

### Step 6: Final result
Therefore, the value of the integral is:
\[ I = \frac{\pi}{4} \]