`inte^(3 log x ) (x^(4)+ 1) ^(-1) dx` is equal to

Options

A. `(1)/( 4) log ( x^(4) + 1) + C`
B. `- ( 1)/( 4) log ( x^(4) + 1) +C`
C. `log (x^(4) + 1) + C`
D. none of these

Answer

To solve the integral \( \int e^{3 \log x} (x^4 + 1)^{-1} \, dx \), we can follow these steps:

### Step 1: Simplify \( e^{3 \log x} \)
Using the property of logarithms, we know that \( e^{a \log b} = b^a \). Therefore, we can rewrite \( e^{3 \log x} \) as:
\[ e^{3 \log x} = x^3 \]

### Step 2: Rewrite the integral
Now, substituting \( e^{3 \log x} \) in the integral, we have:
\[ \int e^{3 \log x} (x^4 + 1)^{-1} \, dx = \int \frac{x^3}{x^4 + 1} \, dx \]

### Step 3: Use substitution
Let’s use the substitution:
\[ t = x^4 + 1 \]
Then, differentiating both sides gives:
\[ dt = 4x^3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{4x^3} \]

### Step 4: Substitute in the integral
Now we can substitute \( dx \) in the integral:
\[ \int \frac{x^3}{x^4 + 1} \, dx = \int \frac{x^3}{t} \cdot \frac{dt}{4x^3} = \int \frac{1}{4t} \, dt \]

### Step 5: Integrate
Now, we can integrate:
\[ \int \frac{1}{4t} \, dt = \frac{1}{4} \ln |t| + C \]

### Step 6: Substitute back for \( t \)
Now, substituting back for \( t \):
\[ \frac{1}{4} \ln |x^4 + 1| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int e^{3 \log x} (x^4 + 1)^{-1} \, dx = \frac{1}{4} \ln(x^4 + 1) + C \]

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