`l=inttan^(-1)""((1)/(x^(2)-x+1))dx` is equal to

Options

A. `xtan^(-1)((1)/(x^(2)-x+1))+log_(e)""(x)/((2-x)sqrt(1+x^(2))sqrt(2x-x^(2)))+c`
B. `xcot^(-1)(x^(2)-x+1)+log_(e)""(x)/((2-x)sqrt(1+x^(2))sqrt(2x-x^(2)))+c`
C. `tan^(-1)((1)/(x^(2)-x+1))+c`
D. `xtan^(-1)""(1)/(x^(2)-x+1)-tan^(-1)(1-x)+(1)/(2)log""(x^(2)-2x+2)/(x^(2)+1)+c`

Answer

To solve the integral \( I = \int \tan^{-1}\left(\frac{1}{x^2 - x + 1}\right) dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rearranging the Integral**:
We start with the integral:
\[ I = \int \tan^{-1}\left(\frac{1}{x^2 - x + 1}\right) dx \]
We can rewrite the denominator:
\[ x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4} \]
However, for simplicity, we will proceed with the original form.

2. **Using the Formula for Inverse Tangent**:
Recall the formula:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]
We can express our integral as:
\[ I = \int \left( \tan^{-1}(x) + \tan^{-1}(1 - x) \right) dx \]

3. **Splitting the Integral**:
We can split the integral into two parts:
\[ I = \int \tan^{-1}(x) dx + \int \tan^{-1}(1 - x) dx \]
Let’s denote:
\[ I_1 = \int \tan^{-1}(x) dx \quad \text{and} \quad I_2 = \int \tan^{-1}(1 - x) dx \]

4. **Solving \( I_1 \) using Integration by Parts**:
For \( I_1 \), we will use integration by parts:
\[ u = \tan^{-1}(x), \quad dv = dx \]
Then,
\[ du = \frac{1}{1 + x^2} dx, \quad v = x \]
Applying integration by parts:
\[ I_1 = x \tan^{-1}(x) - \int \frac{x}{1 + x^2} dx \]

5. **Evaluating the Remaining Integral**:
The integral \( \int \frac{x}{1 + x^2} dx \) can be solved by substitution:
Let \( w = 1 + x^2 \), then \( dw = 2x dx \) or \( dx = \frac{dw}{2x} \).
Thus,
\[ \int \frac{x}{1 + x^2} dx = \frac{1}{2} \ln|1 + x^2| + C \]
Therefore,
\[ I_1 = x \tan^{-1}(x) - \frac{1}{2} \ln|1 + x^2| + C \]

6. **Solving \( I_2 \)**:
For \( I_2 \), we can use a similar approach:
Let \( t = 1 - x \), then \( dt = -dx \). Thus,
\[ I_2 = -\int \tan^{-1}(t) dt \]
This will yield:
\[ I_2 = -\left( t \tan^{-1}(t) - \frac{1}{2} \ln|1 + t^2| \right) + C \]
Substituting back \( t = 1 - x \):
\[ I_2 = -(1 - x) \tan^{-1}(1 - x) + \frac{1}{2} \ln|1 + (1 - x)^2| + C \]

7. **Combining \( I_1 \) and \( I_2 \)**:
Finally, we combine both integrals:
\[ I = I_1 + I_2 \]
After simplification, we can express the final answer in terms of \( x \).

### Final Result:
The final result of the integral is:
\[ I = x \tan^{-1}(x) - (1 - x) \tan^{-1}(1 - x) - \frac{1}{2} \ln|1 + x^2| + \frac{1}{2} \ln|1 + (1 - x)^2| + C \]