The value of `int_(1/e->tanx) (tdt)/(1+t^2) + int_(1/e->cotx) (dt)/(t*(1+t^2)) =`

Options

A. (a) 0
B. (b) 2
C. (c) 1
D. (c) none of these

Explanation

To solve the given problem, we need to evaluate the following expression:

\[ \int_{1/e}^{\tan x} \frac{t \, dt}{1 + t^2} + \int_{1/e}^{\cot x} \frac{dt}{t(1 + t^2)} \]

### Step 1: Evaluate the first integral

We start with the first integral:

\[ I_1 = \int_{1/e}^{\tan x} \frac{t \, dt}{1 + t^2} \]

To solve this integral, we can use the substitution \( p = 1 + t^2 \). Then, differentiating gives:

\[ dp = 2t \, dt \quad \Rightarrow \quad dt = \frac{dp}{2t} \]

Substituting \( t = \sqrt{p - 1} \) into the integral, we have:

\[ I_1 = \int_{1/e}^{\tan x} \frac{t \cdot \frac{dp}{2t}}{p} = \frac{1}{2} \int_{1 + (1/e)^2}^{1 + \tan^2 x} \frac{dp}{p} \]

The limits change as follows:
- When \( t = 1/e \), \( p = 1 + (1/e)^2 = 1 + 1/e^2 \)
- When \( t = \tan x \), \( p = 1 + \tan^2 x = \sec^2 x \)

Thus, we can rewrite the integral:

\[ I_1 = \frac{1}{2} \left[ \log p \right]_{1 + 1/e^2}^{\sec^2 x} = \frac{1}{2} \left( \log(\sec^2 x) - \log(1 + 1/e^2) \right) \]

Using the property of logarithms, we can simplify this to:

\[ I_1 = \frac{1}{2} \left( \log\left(\frac{\sec^2 x}{1 + 1/e^2}\right) \right) \]

### Step 2: Evaluate the second integral

Now we evaluate the second integral:

\[ I_2 = \int_{1/e}^{\cot x} \frac{dt}{t(1 + t^2)} \]

This integral can be split using the property of logarithms:

\[ I_2 = \int_{1/e}^{\cot x} \left( \frac{1}{t} - \frac{t}{1 + t^2} \right) dt \]

Calculating each part separately:

1. The first part:

\[ \int_{1/e}^{\cot x} \frac{1}{t} dt = \left[ \log t \right]_{1/e}^{\cot x} = \log(\cot x) - \log(1/e) = \log(\cot x) + 1 \]

2. The second part:

\[ \int_{1/e}^{\cot x} \frac{t}{1 + t^2} dt \]

Using the same substitution \( p = 1 + t^2 \):

\[ = \frac{1}{2} \left[ \log(1 + t^2) \right]_{1/e}^{\cot x} = \frac{1}{2} \left( \log(1 + \cot^2 x) - \log(1 + 1/e^2) \right) \]

Since \( 1 + \cot^2 x = \csc^2 x \), we have:

\[ = \frac{1}{2} \left( \log(\csc^2 x) - \log(1 + 1/e^2) \right) \]

### Step 3: Combine the results

Now, we can combine \( I_1 \) and \( I_2 \):

\[ I = I_1 + I_2 = \frac{1}{2} \left( \log\left(\frac{\sec^2 x}{1 + 1/e^2}\right) \right) + \left( \log(\cot x) + 1 - \frac{1}{2} \left( \log(\csc^2 x) - \log(1 + 1/e^2) \right) \right) \]

This simplifies to:

\[ I = \frac{1}{2} \log\left(\frac{\sec^2 x}{1 + 1/e^2}\right) + \log(\cot x) + 1 - \frac{1}{2} \log(\csc^2 x) + \frac{1}{2} \log(1 + 1/e^2) \]

### Step 4: Final simplification

After simplifying the logarithmic terms, we find that the expression reduces to:

\[ I = 1 \]

### Conclusion

Thus, the final answer is:

\[ \boxed{1} \]