The value of `sum_(m=1)^ootan^(- 1)((2m)/(m^4+m^2+2))` is

Answer

To solve the problem, we need to evaluate the infinite sum:

\[ \sum_{m=1}^{\infty} \tan^{-1}\left(\frac{2m}{m^4 + m^2 + 2}\right) \]

### Step 1: Simplify the Argument of the Inverse Tangent
We start with the expression inside the inverse tangent function:

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

We can rewrite the denominator:

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

### Step 2: Factor the Denominator
We can express the denominator as:

\[ (m^2 + 1)^2 + 1 - m^2 = (m^2 + 1 - m)(m^2 + 1 + m) \]

This gives us:

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

### Step 3: Use the Identity for Inverse Tangent
Using the identity:

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

we can express our sum as:

\[ \tan^{-1}(m^2 + m + 1) - \tan^{-1}(m^2 + 1 - m) \]

### Step 4: Write the Infinite Sum
Now we can write the infinite sum:

\[ \sum_{m=1}^{\infty} \left( \tan^{-1}(m^2 + m + 1) - \tan^{-1}(m^2 + 1 - m) \right) \]

### Step 5: Identify the Telescoping Nature
Notice that this sum is telescoping. When we expand it, we have:

- For \( m=1 \): \( \tan^{-1}(3) - \tan^{-1}(1) \)
- For \( m=2 \): \( \tan^{-1}(7) - \tan^{-1}(3) \)
- For \( m=3 \): \( \tan^{-1}(13) - \tan^{-1}(7) \)
- And so on...

### Step 6: Evaluate the Limit
As \( m \) approaches infinity, \( \tan^{-1}(m^2 + m + 1) \) approaches \( \frac{\pi}{2} \) and \( \tan^{-1}(1) \) is \( \frac{\pi}{4} \).

Thus, the sum converges to:

\[ \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Final Answer
The value of the infinite sum is:

\[ \frac{\pi}{4} \]