`int(x^(4)+1)/(x^(6)+1)dx=`

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

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{x^4 + 1}{x^6 + 1} \, dx \]
We can factor the denominator \( x^6 + 1 \) as \( (x^2 + 1)(x^4 - x^2 + 1) \).

### Step 2: Split the Fraction
Next, we can express the integrand in a form that allows us to separate it:
\[ \frac{x^4 + 1}{x^6 + 1} = \frac{x^4 + 1 - x^2 + x^2}{x^6 + 1} = \frac{(x^4 - x^2 + 1) + x^2}{x^6 + 1} \]
This gives us:
\[ \int \left( \frac{x^4 - x^2 + 1}{x^6 + 1} + \frac{x^2}{x^6 + 1} \right) \, dx \]

### Step 3: Simplify the Integral
Now we can split the integral:
\[ \int \frac{x^4 - x^2 + 1}{x^6 + 1} \, dx + \int \frac{x^2}{x^6 + 1} \, dx \]

### Step 4: Use Substitution for the Second Integral
For the second integral, we can use the substitution \( t = x^3 \), which implies \( dt = 3x^2 \, dx \) or \( dx = \frac{dt}{3x^2} \). Thus:
\[ \int \frac{x^2}{x^6 + 1} \, dx = \frac{1}{3} \int \frac{1}{t^2 + 1} \, dt \]
The integral \( \int \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t) + C \), so we have:
\[ \frac{1}{3} \tan^{-1}(x^3) + C \]

### Step 5: Solve the First Integral
The first integral can be simplified further. We can notice that:
\[ \int \frac{x^4 - x^2 + 1}{x^6 + 1} \, dx \]
is a bit complicated, but we can use polynomial long division or partial fractions if necessary. However, it can be shown that:
\[ \int \frac{x^4 + 1}{x^6 + 1} \, dx = \tan^{-1}(x) + \frac{1}{3} \tan^{-1}(x^3) + C \]

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