`intsqrt((e^x+1)/(e^x-1))dx` `(A) ln (e^(x)+sqrt(e^(2x)-1))-sec^(-1)(e^(x))`+C `(B) ln(e^(x)+sqrt(e^(2x)-1))+sec^(-1)(e^(x))+C` `(C) ln (e^(x)-sqrt(e^(2x)-1))-sec^(-1)(e^(x))`+C `(D) ln(e^(x)+sqrt(e^(2x)-1))-sin^(-1)(e^(-x))+C`

Options

A. `ln (e^(x)+sqrt(e^(2x)-1))-sec^(-1)(e^(x))`+C
B. `ln(e^(x)+sqrt(e^(2x)-1)+sec^(-1)(e^(x))+C`
C. `ln (e^(x)-sqrt(e^(2x)-1)-sec^(-1)(e^(x))`+C
D. `ln(e^(x)+sqrt(e^(2x)-1)-sin^(-1)(e^(x))+C`

Answer

To solve the integral \( \int \sqrt{\frac{e^x + 1}{e^x - 1}} \, dx \), we can follow these steps:

### Step 1: Rationalize the Integral
Multiply and divide the integrand by \( e^x + 1 \):

\[ \int \sqrt{\frac{e^x + 1}{e^x - 1}} \, dx = \int \frac{(e^x + 1) \sqrt{e^x + 1}}{\sqrt{(e^x - 1)(e^x + 1)}} \, dx \]

### Step 2: Simplify the Denominator
The denominator can be simplified using the difference of squares:

\[ \sqrt{(e^x - 1)(e^x + 1)} = \sqrt{e^{2x} - 1} \]

Thus, the integral becomes:

\[ \int \frac{(e^x + 1)}{\sqrt{e^{2x} - 1}} \, dx \]

### Step 3: Split the Integral
Now, we can split the integral into two parts:

\[ I = \int \frac{e^x}{\sqrt{e^{2x} - 1}} \, dx + \int \frac{1}{\sqrt{e^{2x} - 1}} \, dx \]

Let’s denote these two integrals as \( I_1 \) and \( I_2 \):

\[ I_1 = \int \frac{e^x}{\sqrt{e^{2x} - 1}} \, dx, \quad I_2 = \int \frac{1}{\sqrt{e^{2x} - 1}} \, dx \]

### Step 4: Solve \( I_1 \)
For \( I_1 \), use the substitution \( t = e^x \), which gives \( dt = e^x \, dx \) or \( dx = \frac{dt}{t} \):

\[ I_1 = \int \frac{t}{\sqrt{t^2 - 1}} \cdot \frac{dt}{t} = \int \frac{1}{\sqrt{t^2 - 1}} \, dt \]

The integral \( \int \frac{1}{\sqrt{t^2 - 1}} \, dt \) is known to be:

\[ \ln(t + \sqrt{t^2 - 1}) + C \]

Substituting back \( t = e^x \):

\[ I_1 = \ln(e^x + \sqrt{e^{2x} - 1}) + C_1 \]

### Step 5: Solve \( I_2 \)
For \( I_2 \), we can use the same substitution \( t = e^x \):

\[ I_2 = \int \frac{1}{\sqrt{t^2 - 1}} \cdot \frac{dt}{t} = \int \frac{1}{\sqrt{t^2 - 1}} \, dt \]

This integral is:

\[ \sin^{-1}\left(\frac{1}{t}\right) + C_2 \]

Substituting back \( t = e^x \):

\[ I_2 = -\sin^{-1}(e^{-x}) + C_2 \]

### Step 6: Combine Results
Now combine both integrals:

\[ I = I_1 + I_2 = \ln(e^x + \sqrt{e^{2x} - 1}) - \sin^{-1}(e^{-x}) + C \]

### Final Answer
Thus, the final answer is:

\[ \int \sqrt{\frac{e^x + 1}{e^x - 1}} \, dx = \ln(e^x + \sqrt{e^{2x} - 1}) - \sin^{-1}(e^{-x}) + C \]