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

Answer

To evaluate the integral \( \int \frac{x^4 + 1}{x^6 + 1} \, dx \), we will follow a systematic approach.

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{x^4 + 1}{x^6 + 1} \, dx \]

### Step 2: Multiply by a Suitable Form
To simplify the integrand, we can multiply the numerator and the denominator by \( x^2 + 1 \):
\[ I = \int \frac{(x^4 + 1)(x^2 + 1)}{(x^6 + 1)(x^2 + 1)} \, dx \]

### Step 3: Expand the Numerator
Now, we expand the numerator:
\[ (x^4 + 1)(x^2 + 1) = x^6 + x^4 + x^2 + 1 \]
So, we have:
\[ I = \int \frac{x^6 + x^4 + x^2 + 1}{(x^6 + 1)(x^2 + 1)} \, dx \]

### Step 4: Split the Integral
Now we can split the integral into two parts:
\[ I = \int \frac{1}{x^2 + 1} \, dx + \int \frac{x^2}{x^6 + 1} \, dx \]

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

### Step 6: Evaluate the Second Integral
For the second integral, we can use the substitution \( x^3 = t \), which gives us \( 3x^2 \, dx = dt \) or \( dx = \frac{dt}{3x^2} \):
\[ \int \frac{x^2}{x^6 + 1} \, dx = \int \frac{x^2}{t^2 + 1} \cdot \frac{dt}{3x^2} = \frac{1}{3} \int \frac{1}{t^2 + 1} \, dt \]
This integral evaluates to:
\[ \frac{1}{3} \tan^{-1}(t) + C_2 = \frac{1}{3} \tan^{-1}(x^3) + C_2 \]

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

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