`f(x)=int_1^x(tan^(-1)(t))/t dtAAx in R^+,t h e nfin dt h ev a l u eof` `f(e^2)-f(1/(e^2))`

Explanation

To solve the problem, we need to find the value of \( f(e^2) - f\left(\frac{1}{e^2}\right) \) where

\[ f(x) = \int_1^x \frac{\tan^{-1}(t)}{t} \, dt. \]

### Step-by-step Solution:

1. **Define the function**:
\[ f(x) = \int_1^x \frac{\tan^{-1}(t)}{t} \, dt. \]

2. **Find \( f\left(\frac{1}{x}\right) \)**:
We need to express \( f\left(\frac{1}{x}\right) \):
\[ f\left(\frac{1}{x}\right) = \int_1^{\frac{1}{x}} \frac{\tan^{-1}(t)}{t} \, dt. \]

3. **Change of Variables**:
Let \( t = \frac{1}{u} \). Then \( dt = -\frac{1}{u^2} \, du \).
- When \( t = 1 \), \( u = 1 \).
- When \( t = \frac{1}{x} \), \( u = x \).

Thus, we have:
\[ f\left(\frac{1}{x}\right) = \int_1^{x} \frac{\tan^{-1}\left(\frac{1}{u}\right)}{\frac{1}{u}} \left(-\frac{1}{u^2}\right) \, du. \]
This simplifies to:
\[ f\left(\frac{1}{x}\right) = -\int_1^{x} \frac{\tan^{-1}\left(\frac{1}{u}\right)}{u} \, du. \]

4. **Using the Identity**:
We know that:
\[ \tan^{-1}\left(\frac{1}{u}\right) = \frac{\pi}{2} - \tan^{-1}(u). \]
Thus, we can rewrite:
\[ f\left(\frac{1}{x}\right) = -\int_1^{x} \frac{\frac{\pi}{2} - \tan^{-1}(u)}{u} \, du. \]
This expands to:
\[ f\left(\frac{1}{x}\right) = -\left(\frac{\pi}{2} \int_1^{x} \frac{1}{u} \, du - \int_1^{x} \frac{\tan^{-1}(u)}{u} \, du\right). \]

5. **Calculate the Integrals**:
The integral \( \int_1^{x} \frac{1}{u} \, du = \log(x) \), so:
\[ f\left(\frac{1}{x}\right) = -\left(\frac{\pi}{2} \log(x) - f(x)\right). \]

6. **Combine the Results**:
Thus, we have:
\[ f(x) - f\left(\frac{1}{x}\right) = f(x) + \left(\frac{\pi}{2} \log(x) - f(x)\right) = \frac{\pi}{2} \log(x). \]

7. **Substituting Values**:
Now, substituting \( x = e^2 \):
\[ f(e^2) - f\left(\frac{1}{e^2}\right) = \frac{\pi}{2} \log(e^2) = \frac{\pi}{2} \cdot 2 = \pi. \]

### Final Answer:
\[ f(e^2) - f\left(\frac{1}{e^2}\right) = \pi. \]