Let `x_(1) , x_(2) , x_(3)` be the solution of `tan^(-1) ((2x + 1)/(x +1 )) + tan ^(-1) ((2x - 1)/( x -1 )) = 2 tan ^(-1) ( x + 1) " where " x_(1) lt x_(2) lt x_(3) " , then " 2x_(1) + x_(2) + x_(3)^(2) ` is equal to

Answer

To solve the equation

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

we will use the properties of inverse tangent functions.

### Step 1: Use the formula for the sum of arctangents

We know that

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

provided \( ab < 1 \). Let \( a = \frac{2x + 1}{x + 1} \) and \( b = \frac{2x - 1}{x - 1} \).

### Step 2: Calculate \( a + b \)

\[ a + b = \frac{2x + 1}{x + 1} + \frac{2x - 1}{x - 1} = \frac{(2x + 1)(x - 1) + (2x - 1)(x + 1)}{(x + 1)(x - 1)}. \]

Calculating the numerator:

\[ (2x + 1)(x - 1) + (2x - 1)(x + 1) = (2x^2 - 2x + x - 1) + (2x^2 + 2x - x - 1) = 4x^2 - 2. \]

So,

\[ a + b = \frac{4x^2 - 2}{(x + 1)(x - 1)}. \]

### Step 3: Calculate \( ab \)

\[ ab = \left(\frac{2x + 1}{x + 1}\right)\left(\frac{2x - 1}{x - 1}\right) = \frac{(2x + 1)(2x - 1)}{(x + 1)(x - 1)} = \frac{4x^2 - 1}{x^2 - 1}. \]

### Step 4: Substitute into the equation

Now substituting into the left-hand side:

\[ \tan^{-1}\left(\frac{4x^2 - 2}{(x + 1)(x - 1) - (4x^2 - 1)}\right) = 2\tan^{-1}(x + 1). \]

Using the double angle formula for tangent:

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

### Step 5: Set the two expressions equal

Equating the two expressions gives:

\[ \frac{4x^2 - 2}{(x + 1)(x - 1) - (4x^2 - 1)} = \frac{2(x + 1)}{1 - (x + 1)^2}. \]

### Step 6: Simplify and solve for \( x \)

Cross-multiplying and simplifying leads to a polynomial equation in \( x \):

\[ 2x^3 - 4x^2 - x - 2 = 0. \]

### Step 7: Factor the polynomial

Factoring gives:

\[ (x^2 - 2)(2x + 1) = 0. \]

### Step 8: Find the roots

The roots are:

1. \( x^2 - 2 = 0 \) gives \( x = \sqrt{2}, -\sqrt{2} \).
2. \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \).

Thus, the solutions are \( x_1 = -\sqrt{2}, x_2 = -\frac{1}{2}, x_3 = \sqrt{2} \).

### Step 9: Calculate \( 2x_1 + x_2 + x_3^2 \)

Now we compute:

\[ 2x_1 + x_2 + x_3^2 = 2(-\sqrt{2}) + \left(-\frac{1}{2}\right) + (\sqrt{2})^2. \]

Calculating this gives:

\[ = -2\sqrt{2} - \frac{1}{2} + 2 = -2\sqrt{2} + \frac{3}{2}. \]

### Final Answer

Thus, the final expression is:

\[ 2x_1 + x_2 + x_3^2 = -2\sqrt{2} + \frac{3}{2}. \]