Let `f(x)=(e^(x)+1)/(e^(x)-1) and int_(0)^(1) x^(3) .(e^(x)+1)/(e^(x)-1) dx= alpha "Then" , int_(-1)^(1) t^(3) f(t) dt` is equal to

Options

A. 0
B. `alpha`
C. `2 alpha`
D. None of these

Answer

To solve the problem step by step, we need to evaluate the integral \( \int_{-1}^{1} t^3 f(t) dt \) where \( f(t) = \frac{e^t + 1}{e^t - 1} \).

### Step 1: Define the function
We start with the given function:
\[ f(t) = \frac{e^t + 1}{e^t - 1} \]

### Step 2: Determine the nature of \( t^3 f(t) \)
Next, we need to analyze the function \( g(t) = t^3 f(t) \):
\[ g(t) = t^3 \cdot \frac{e^t + 1}{e^t - 1} \]

### Step 3: Check if \( g(t) \) is even or odd
To check if \( g(t) \) is even or odd, we compute \( g(-t) \):
\[ g(-t) = (-t)^3 \cdot \frac{e^{-t} + 1}{e^{-t} - 1} = -t^3 \cdot \frac{e^{-t} + 1}{e^{-t} - 1} \]
We can simplify \( \frac{e^{-t} + 1}{e^{-t} - 1} \):
\[ \frac{e^{-t} + 1}{e^{-t} - 1} = \frac{1 + e^t}{1 - e^t} = -\frac{1 + e^t}{e^t - 1} \]
Thus,
\[ g(-t) = -t^3 \cdot \left(-\frac{1 + e^t}{e^t - 1}\right) = t^3 \cdot \frac{1 + e^t}{e^t - 1} \]
This shows that:
\[ g(-t) = g(t) \]
indicating that \( g(t) \) is an even function.

### Step 4: Evaluate the integral
Since \( g(t) \) is even, we can use the property of even functions in integrals:
\[ \int_{-1}^{1} g(t) dt = 2 \int_{0}^{1} g(t) dt \]
Thus,
\[ \int_{-1}^{1} t^3 f(t) dt = 2 \int_{0}^{1} t^3 f(t) dt \]

### Step 5: Substitute the value of \( \int_{0}^{1} t^3 f(t) dt \)
From the problem statement, we know:
\[ \int_{0}^{1} t^3 f(t) dt = \alpha \]
Therefore,
\[ \int_{-1}^{1} t^3 f(t) dt = 2 \alpha \]

### Final Answer
The final result is:
\[ \int_{-1}^{1} t^3 f(t) dt = 2\alpha \]