`int(x^3-x)/(1+x^6)dx` is equal to

Options

A. `(1)/(6)log.(x^(4)-x^(2)+1)/(x(x^(2)+1))+C`
B. `(1)/(6)tan^(-1).((x^(2)+1)^(2))/(2)+C`
C. `log.(x^(4)-x^(2)+1)/((1+x^(2))^(2))+C`
D. `tan^(-1).((x^(2)+1)^(2))/(2)+C`

Explanation

To solve the integral \( \int \frac{x^3 - x}{1 + x^6} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
We can factor out \( x \) from the numerator:
\[ \frac{x^3 - x}{1 + x^6} = \frac{x(x^2 - 1)}{1 + x^6} \]

### Step 2: Substitute \( x^2 = t \)
Let \( t = x^2 \). Then, \( dx = \frac{dt}{2\sqrt{t}} \) or \( 2x \, dx = dt \). Thus, \( x = \sqrt{t} \) and \( x^3 = t\sqrt{t} \). The integral becomes:
\[ \int \frac{\sqrt{t}(t - 1)}{1 + t^3} \cdot \frac{dt}{2\sqrt{t}} = \frac{1}{2} \int \frac{t - 1}{1 + t^3} \, dt \]

### Step 3: Split the integral
Now we can split the integral:
\[ \frac{1}{2} \int \frac{t - 1}{1 + t^3} \, dt = \frac{1}{2} \left( \int \frac{t}{1 + t^3} \, dt - \int \frac{1}{1 + t^3} \, dt \right) \]

### Step 4: Integrate \( \int \frac{t}{1 + t^3} \, dt \)
For the first integral, we can use the substitution \( u = 1 + t^3 \), which gives \( du = 3t^2 \, dt \) or \( dt = \frac{du}{3t^2} \). Thus,
\[ \int \frac{t}{1 + t^3} \, dt = \frac{1}{3} \int \frac{1}{u} \, du = \frac{1}{3} \ln |1 + t^3| + C \]

### Step 5: Integrate \( \int \frac{1}{1 + t^3} \, dt \)
For the second integral, we can use partial fractions:
\[ \frac{1}{1 + t^3} = \frac{1}{(t + 1)(t^2 - t + 1)} \]
This can be decomposed into partial fractions:
\[ \frac{A}{t + 1} + \frac{Bt + C}{t^2 - t + 1} \]
Solving for \( A, B, C \) gives us the coefficients needed to integrate.

### Step 6: Combine the results
After integrating both parts, we combine the results and substitute back \( t = x^2 \) to express everything in terms of \( x \).

### Final Answer
The final expression will be:
\[ \frac{1}{6} \ln(1 + x^4) - \frac{1}{6} \ln(x^4 - x^2 + 1) + C \]