`int (x)/((1-x^(4)))dx=?`

Options

A. `(1)/(4)log |(1+x^(2))/(1-x^(2))|+C`
B. `(1)/(4)log |(1-x^(2))/(1+x^(2))|+C`
C. `(1)/(2)log |(1+x^(2))/(1-x^(2))|+C`
D. None of these

Explanation

To solve the integral \( \int \frac{x}{1 - x^4} \, dx \), we will use the method of substitution and partial fractions. Here’s a step-by-step solution:

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

### Step 2: Factor the Denominator
Notice that \( 1 - x^4 \) can be factored as:
\[ 1 - x^4 = (1 - x^2)(1 + x^2) \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{x}{(1 - x^2)(1 + x^2)} \, dx \]

### Step 3: Use Substitution
Let \( t = x^2 \). Then, we have:
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} = \frac{dt}{2\sqrt{t}} \]
Now, substituting \( x^2 = t \) into the integral, we get:
\[ I = \int \frac{\sqrt{t}}{(1 - t)(1 + t)} \cdot \frac{dt}{2\sqrt{t}} = \frac{1}{2} \int \frac{1}{(1 - t)(1 + t)} \, dt \]

### Step 4: Partial Fraction Decomposition
Next, we perform partial fraction decomposition on \( \frac{1}{(1 - t)(1 + t)} \):
\[ \frac{1}{(1 - t)(1 + t)} = \frac{A}{1 - t} + \frac{B}{1 + t} \]
Multiplying through by the denominator \( (1 - t)(1 + t) \) gives:
\[ 1 = A(1 + t) + B(1 - t) \]
Setting \( t = 1 \):
\[ 1 = A(2) \quad \Rightarrow \quad A = \frac{1}{2} \]
Setting \( t = -1 \):
\[ 1 = B(2) \quad \Rightarrow \quad B = \frac{1}{2} \]
Thus, we have:
\[ \frac{1}{(1 - t)(1 + t)} = \frac{1/2}{1 - t} + \frac{1/2}{1 + t} \]

### Step 5: Integrate Each Term
Now substituting back into the integral:
\[ I = \frac{1}{2} \int \left( \frac{1/2}{1 - t} + \frac{1/2}{1 + t} \right) dt \]
This simplifies to:
\[ I = \frac{1}{4} \int \frac{1}{1 - t} \, dt + \frac{1}{4} \int \frac{1}{1 + t} \, dt \]
Integrating each term gives:
\[ I = \frac{1}{4} \left( -\log|1 - t| + \log|1 + t| \right) + C \]

### Step 6: Substitute Back
Now we substitute back \( t = x^2 \):
\[ I = \frac{1}{4} \left( -\log|1 - x^2| + \log|1 + x^2| \right) + C \]
This can be combined using properties of logarithms:
\[ I = \frac{1}{4} \log \left| \frac{1 + x^2}{1 - x^2} \right| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{x}{1 - x^4} \, dx = \frac{1}{4} \log \left| \frac{1 + x^2}{1 - x^2} \right| + C \]