`int(x^(4)+1)/(x^(6)+1)dx` is equal to

Options

A. `tan^(-1)x+(1)/(3)tan^(-1)x^(3)+C`
B. `tan^(-1)x-(1)/(3)tan^(-1)x^(3)+C`
C. `-tan^(-1)x-(1)/(3)tan^(-1)x^(3)+C`
D. none of these

Explanation

To solve the integral \( I = \int \frac{x^4 + 1}{x^6 + 1} \, dx \), we can follow these steps:

### Step 1: Rewrite the Denominator
We can express the denominator \( x^6 + 1 \) as \( (x^2)^3 + 1^3 \). This allows us to use the sum of cubes factorization:
\[ x^6 + 1 = (x^2 + 1)((x^2)^2 - x^2 \cdot 1 + 1^2) = (x^2 + 1)(x^4 - x^2 + 1) \]

### Step 2: Set Up the Integral
Now, we rewrite the integral:
\[ I = \int \frac{x^4 + 1}{(x^2 + 1)(x^4 - x^2 + 1)} \, dx \]

### Step 3: Perform Polynomial Long Division
Since the degree of the numerator is less than the degree of the denominator, we can directly proceed to separate the fraction:
\[ I = \int \left( \frac{x^4 - x^2 + 1}{(x^2 + 1)(x^4 - x^2 + 1)} + \frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \right) \, dx \]

### Step 4: Split the Integral
Now we split the integral into two parts:
\[ I = \int \frac{1}{x^2 + 1} \, dx + \int \frac{x^2}{x^4 - x^2 + 1} \, dx \]

### Step 5: Solve the First Integral
The first integral is straightforward:
\[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) + C_1 \]

### Step 6: Solve the Second Integral
For the second integral, we can use the substitution \( t = x^3 \), then \( dt = 3x^2 \, dx \) or \( dx = \frac{dt}{3x^2} \):
\[ \int \frac{x^2}{x^4 - x^2 + 1} \, dx = \frac{1}{3} \int \frac{1}{t^2 + 1} \, dt = \frac{1}{3} \tan^{-1}(t) + C_2 \]
Substituting back \( t = x^3 \):
\[ \frac{1}{3} \tan^{-1}(x^3) + C_2 \]

### Step 7: Combine the Results
Now we combine the results from both integrals:
\[ I = \tan^{-1}(x) + \frac{1}{3} \tan^{-1}(x^3) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{x^4 + 1}{x^6 + 1} \, dx = \tan^{-1}(x) + \frac{1}{3} \tan^{-1}(x^3) + C \]