Solve for x , `tan^(-1) ( x + 1) + tan^(-1) x + tan^(-1) ( x - 1) = tan ^(-1) 3x`

Answer

To solve the equation \( \tan^{-1}(x + 1) + \tan^{-1}(x) + \tan^{-1}(x - 1) = \tan^{-1}(3x) \), we will use the properties of inverse tangent functions.

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate one of the inverse tangent terms:
\[ \tan^{-1}(x + 1) + \tan^{-1}(x - 1) = \tan^{-1}(3x) - \tan^{-1}(x) \]

### Step 2: Using the Identity for Addition of Inverse Tangents
We use the identity for the sum of two inverse tangents:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
Applying this to the left-hand side:
\[ \tan^{-1}(x + 1) + \tan^{-1}(x - 1) = \tan^{-1}\left(\frac{(x + 1) + (x - 1)}{1 - (x + 1)(x - 1)}\right) \]
This simplifies to:
\[ \tan^{-1}\left(\frac{2x}{1 - (x^2 - 1)}\right) = \tan^{-1}\left(\frac{2x}{2 - x^2}\right) \]

### Step 3: Simplifying the Right-Hand Side
Now we simplify the right-hand side using the identity for the difference of inverse tangents:
\[ \tan^{-1}(3x) - \tan^{-1}(x) = \tan^{-1}\left(\frac{3x - x}{1 + 3x^2}\right) = \tan^{-1}\left(\frac{2x}{1 + 3x^2}\right) \]

### Step 4: Setting the Arguments Equal
Since both sides are equal, we can set their arguments equal:
\[ \frac{2x}{2 - x^2} = \frac{2x}{1 + 3x^2} \]

### Step 5: Cross Multiplying
Cross multiplying gives us:
\[ 2x(1 + 3x^2) = 2x(2 - x^2) \]
Assuming \( x \neq 0 \) (we will check for \( x = 0 \) later), we can divide both sides by \( 2x \):
\[ 1 + 3x^2 = 2 - x^2 \]

### Step 6: Rearranging the Equation
Rearranging the equation leads to:
\[ 3x^2 + x^2 = 2 - 1 \]
\[ 4x^2 = 1 \]

### Step 7: Solving for \( x \)
Dividing both sides by 4 gives:
\[ x^2 = \frac{1}{4} \]
Taking the square root gives:
\[ x = \pm \frac{1}{2} \]

### Step 8: Checking for \( x = 0 \)
Now we check if \( x = 0 \) is a solution:
Substituting \( x = 0 \) into the original equation:
\[ \tan^{-1}(1) + \tan^{-1}(0) + \tan^{-1}(-1) = \tan^{-1}(0) \]
This simplifies to:
\[ \frac{\pi}{4} + 0 - \frac{\pi}{4} = 0 \]
This is true, so \( x = 0 \) is also a solution.

### Final Solutions
Thus, the solutions for the equation are:
\[ x = \frac{1}{2}, \quad x = -\frac{1}{2}, \quad x = 0 \]