The value of `int_(0)^(1) tan^(-1)((2x-1)/(1+x-x^(2)))dx` is

Options

A. 1
B. 0
C. -1
D. 2

Answer

To solve the integral \( I = \int_0^1 \tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx \), we can use the properties of definite integrals and the symmetry of the function involved.

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int_0^1 \tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx \]

### Step 2: Use the Substitution \( x = 1 - t \)
Let’s perform the substitution \( x = 1 - t \). Then, when \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 0 \). The differential \( dx \) becomes \( -dt \). Thus, we can rewrite the integral as:
\[ I = \int_1^0 \tan^{-1}\left(\frac{2(1-t)-1}{1+(1-t)-(1-t)^2}\right)(-dt) \]
This simplifies to:
\[ I = \int_0^1 \tan^{-1}\left(\frac{2 - 2t - 1}{1 + 1 - t - (1 - 2t + t^2)}\right)dt \]

### Step 3: Simplify the Argument of \( \tan^{-1} \)
Now, simplify the argument:
\[ \frac{1 - 2t}{1 + 1 - t - (1 - 2t + t^2)} = \frac{1 - 2t}{1 + 1 - t - 1 + 2t - t^2} = \frac{1 - 2t}{1 + t - t^2} \]
Thus, we have:
\[ I = \int_0^1 \tan^{-1}\left(\frac{1 - 2t}{1 + t - t^2}\right)dt \]

### Step 4: Combine the Two Integrals
Now we have two expressions for \( I \):
1. \( I = \int_0^1 \tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx \)
2. \( I = \int_0^1 \tan^{-1}\left(\frac{1 - 2x}{1 + x - x^2}\right)dx \)

Adding these two integrals:
\[ 2I = \int_0^1 \left( \tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right) + \tan^{-1}\left(\frac{1 - 2x}{1 + x - x^2}\right) \right) dx \]

### Step 5: Use the Identity for \( \tan^{-1} \)
Using the identity \( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \):
Let \( a = \frac{2x-1}{1+x-x^2} \) and \( b = \frac{1-2x}{1+x-x^2} \):
\[ a + b = \frac{(2x-1) + (1-2x)}{1+x-x^2} = \frac{0}{1+x-x^2} = 0 \]
Thus,
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}(0) = 0 \]
Therefore,
\[ 2I = \int_0^1 0 \, dx = 0 \]
This implies that:
\[ I = 0 \]

### Final Answer
The value of the integral is:
\[ \boxed{0} \]