Let `f(x)=int(dx)/(e^(x)+8e^(-x)+4e^(-3x)),g(x)=int(dx)/(e^(3x)+8e^(x)+4e^(-x)).`
`int(f(x)-2g(x))dx`

Options

A. `(1)/(4sqrt3)log|(e^(x)-2e^(-x)-2sqrt3)/(e^(x)+2e^(-x)+2sqrt3)|+C`
B. `(1)/(2)tan^(-1)((e^(x)+2e^(-x))/(2))+C`
C. `(1)/(2sqrt3)tan^(-1)((e^(x)-2e^(-x))/(2sqrt3))+C`
D. `(1)/(4)log|(e^(x)+2e^(-x)-2)/(e^(x)+2e^(-x)+2)|+C`

Explanation

To solve the problem, we need to compute the integral of \( f(x) - 2g(x) \) where:

\[ f(x) = \int \frac{dx}{e^x + 8e^{-x} + 4e^{-3x}} \]
\[ g(x) = \int \frac{dx}{e^{3x} + 8e^x + 4e^{-x}} \]

We need to find:

\[ \int (f(x) - 2g(x)) \, dx \]

### Step 1: Rewrite \( f(x) \) and \( g(x) \)

First, we rewrite \( f(x) \) and \( g(x) \) to have a common denominator.

For \( f(x) \):
\[ f(x) = \int \frac{dx}{e^x + 8e^{-x} + 4e^{-3x}} = \int \frac{e^{2x} \, dx}{e^{3x} + 8e^x + 4} \]

For \( g(x) \):
\[ g(x) = \int \frac{dx}{e^{3x} + 8e^x + 4e^{-x}} = \int \frac{e^{2x} \, dx}{e^{4x} + 8e^{2x} + 4} \]

### Step 2: Combine \( f(x) \) and \( g(x) \)

Now, we can express \( f(x) - 2g(x) \):

\[ f(x) - 2g(x) = \int \frac{e^{2x} \, dx}{e^{3x} + 8e^x + 4} - 2 \int \frac{e^{2x} \, dx}{e^{4x} + 8e^{2x} + 4} \]

### Step 3: Find a common denominator

To combine these integrals, we need a common denominator:

\[ = \int \left( \frac{e^{2x}}{e^{3x} + 8e^x + 4} - 2 \cdot \frac{e^{2x}}{e^{4x} + 8e^{2x} + 4} \right) dx \]

### Step 4: Simplify the expression

Now we can simplify the expression:

\[ = \int \frac{e^{2x}(e^{4x} + 8e^{2x} + 4) - 2e^{2x}(e^{3x} + 8e^x + 4)}{(e^{3x} + 8e^x + 4)(e^{4x} + 8e^{2x} + 4)} \, dx \]

### Step 5: Factor and simplify

After simplifying the numerator, we can factor out common terms and simplify further.

### Step 6: Change of variables

Let \( t = e^x \), then \( dt = e^x \, dx \) or \( dx = \frac{dt}{t} \). Substitute \( t \) into the integral.

### Step 7: Solve the integral

Now we can rewrite the integral in terms of \( t \) and solve it using standard integration techniques.

### Step 8: Back-substitute

After finding the integral in terms of \( t \), we will back-substitute \( t = e^x \) to express the final answer in terms of \( x \).

### Final Answer

The final answer will be in the form:

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