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

Options

A. `-1`
B. 0
C. 1
D. `pm1`

Answer

To solve the definite integral

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

we can use the property of definite integrals which states:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In this case, \( a = 0 \) and \( b = 1 \), so we have:

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

Now, let's simplify the expression inside the arctangent function:

1. **Calculate \( 2(1-x) - 1 \)**:
\[ 2(1-x) - 1 = 2 - 2x - 1 = 1 - 2x. \]

2. **Calculate \( 1 + (1-x) - (1-x)^2 \)**:
\[ 1 + (1-x) - (1-x)^2 = 1 + 1 - x - (1 - 2x + x^2) = 2 - x - 1 + 2x - x^2 = 1 + x - x^2. \]

Thus, we can rewrite the integral:

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

Now we have:

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

Next, we can combine the two expressions for \( I \):

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

Using the property of the arctangent:

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

we can find:

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

Thus, we have:

\[ 2I = 0 \implies I = 0. \]

Therefore, the value of the definite integral is:

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

### Final Answer:
\[ \int_{0}^{1} \tan^{-1} \left( \frac{2x-1}{1+x-x^2} \right) dx = 0. \]