The number of integral values of x satisfying the equation `tan^(-1) (3x) + tan^(-1) (5x) = tan^(-1) (7x) + tan^(-1) (2x)` is ____

Explanation

To solve the equation \( \tan^{-1}(3x) + \tan^{-1}(5x) = \tan^{-1}(7x) + \tan^{-1}(2x) \), we will use the properties of inverse tangent functions.

### Step-by-Step Solution:

1. **Write the given equation:**
\[ \tan^{-1}(3x) + \tan^{-1}(5x) = \tan^{-1}(7x) + \tan^{-1}(2x) \]

2. **Rearrange the equation:**
\[ \tan^{-1}(3x) - \tan^{-1}(2x) = \tan^{-1}(7x) - \tan^{-1}(5x) \]

3. **Apply the formula for the difference of inverse tangents:**
The formula states that:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
Applying this to both sides:
\[ \tan^{-1}\left(\frac{3x - 2x}{1 + 3x \cdot 2x}\right) = \tan^{-1}\left(\frac{7x - 5x}{1 + 7x \cdot 5x}\right) \]

4. **Simplify both sides:**
The left side becomes:
\[ \tan^{-1}\left(\frac{x}{1 + 6x^2}\right) \]
The right side becomes:
\[ \tan^{-1}\left(\frac{2x}{1 + 35x^2}\right) \]

5. **Set the arguments equal to each other:**
Since \( \tan^{-1}(a) = \tan^{-1}(b) \) implies \( a = b \) (for \( a, b \) in the range of the tangent function):
\[ \frac{x}{1 + 6x^2} = \frac{2x}{1 + 35x^2} \]

6. **Cross-multiply to eliminate the fractions:**
\[ x(1 + 35x^2) = 2x(1 + 6x^2) \]

7. **Expand both sides:**
\[ x + 35x^3 = 2x + 12x^3 \]

8. **Rearrange the equation:**
\[ 35x^3 - 12x^3 - 2x + x = 0 \]
This simplifies to:
\[ 23x^3 - x = 0 \]

9. **Factor out \( x \):**
\[ x(23x^2 - 1) = 0 \]

10. **Set each factor to zero:**
- \( x = 0 \)
- \( 23x^2 - 1 = 0 \) leads to \( 23x^2 = 1 \) or \( x^2 = \frac{1}{23} \)
- Thus, \( x = \frac{1}{\sqrt{23}} \) and \( x = -\frac{1}{\sqrt{23}} \)

11. **Determine the integral values of \( x \):**
The solutions we found are:
- \( x = 0 \)
- \( x = \frac{1}{\sqrt{23}} \) (not an integer)
- \( x = -\frac{1}{\sqrt{23}} \) (not an integer)

Therefore, the only integral value of \( x \) satisfying the equation is:
\[ \boxed{1} \]