`int e^(3 log x) . (x^4 + 1)^(-1) dx =`

Options

A. `"log" (x^4 + 1)`
B. `3log (x^4 + 1)`
C. `-log (x^4 + 1)`
D. `1/4 "log" (x^4 + 1)`

Answer

To solve the integral

\[ \int e^{3 \log x} \cdot (x^4 + 1)^{-1} \, dx, \]

we will follow these steps:

### Step 1: Simplify the integrand
Using the property of logarithms, we can rewrite \( e^{3 \log x} \) as follows:

\[ e^{3 \log x} = e^{\log(x^3)} = x^3. \]

So, we can rewrite the integral as:

\[ \int x^3 \cdot (x^4 + 1)^{-1} \, dx. \]

### Step 2: Rewrite the integral
Now, we can express the integral in a simpler form:

\[ \int \frac{x^3}{x^4 + 1} \, dx. \]

### Step 3: Use substitution
Let’s use the substitution \( t = x^4 + 1 \). Then, we differentiate \( t \):

\[ dt = 4x^3 \, dx \implies dx = \frac{dt}{4x^3}. \]

### Step 4: Substitute into the integral
Now, substituting \( t \) and \( dx \) into the integral, we have:

\[ \int \frac{x^3}{t} \cdot \frac{dt}{4x^3} = \int \frac{1}{4t} \, dt. \]

### 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, we substitute back \( t = x^4 + 1 \):

\[ \frac{1}{4} \ln |x^4 + 1| + C. \]

### Final Answer
Thus, the final result of the integral is:

\[ \int e^{3 \log x} \cdot (x^4 + 1)^{-1} \, dx = \frac{1}{4} \ln(x^4 + 1) + C. \]

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