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

Options

A. `log(x^4+1)+C`
B. `1/4log(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 will follow these steps:

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

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

### Step 3: Use substitution
Let \( t = x^4 + 1 \). Then, we differentiate both sides:
\[ dt = 4x^3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{4x^3} \]

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

### Step 5: Integrate
The integral of \( \frac{1}{4t} \) is:
\[ \frac{1}{4} \ln |t| + C \]

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

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