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

Options

A. `1//4`
B. `1//2`
C. `1`
D. `2`

Explanation

To solve the integral

\[ I = \int_0^1 \frac{\tan^{-1}\left(\frac{x}{x+1}\right)}{\tan^{-1}\left(\frac{1 + 2x - 2x^2}{2}\right)} \, dx, \]

we can utilize the property of definite integrals that states:

\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx. \]

In this case, \( a = 1 \). Therefore, we can express \( I \) as:

\[ I = \int_0^1 \frac{\tan^{-1}\left(\frac{1-x}{(1-x)+1}\right)}{\tan^{-1}\left(\frac{1 + 2(1-x) - 2(1-x)^2}{2}\right)} \, dx. \]

Now, simplifying the expressions:

1. The first term becomes:
\[ \tan^{-1}\left(\frac{1-x}{2-x}\right). \]

2. The second term simplifies as follows:
\[ 1 + 2(1-x) - 2(1-x)^2 = 1 + 2 - 2x - 2(1 - 2x + x^2) = 1 + 2 - 2x - 2 + 4x - 2x^2 = 2x - 2x^2 + 1. \]
Dividing by 2 gives:
\[ \tan^{-1}\left(\frac{2x - 2x^2 + 1}{2}\right). \]

Thus, we can rewrite \( I \) as:

\[ I = \int_0^1 \frac{\tan^{-1}\left(\frac{1-x}{2-x}\right)}{\tan^{-1}\left(\frac{1 + 2x - 2x^2}{2}\right)} \, dx. \]

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

\[ 2I = \int_0^1 \left( \frac{\tan^{-1}\left(\frac{x}{x+1}\right) + \tan^{-1}\left(\frac{1-x}{2-x}\right)}{\tan^{-1}\left(\frac{1 + 2x - 2x^2}{2}\right)} \right) \, dx. \]

Using the identity for the sum of arctangents:

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

we can simplify the numerator:

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

After simplification, we find that:

\[ 2I = \int_0^1 1 \, dx = 1. \]

Thus, we have:

\[ I = \frac{1}{2}. \]

### Final Answer:
\[ I = \frac{1}{2}. \]