`int(x^2)/(1-x^4)dx`

Explanation

To solve the integral \( I = \int \frac{x^2}{1 - x^4} \, dx \), we will follow these steps:

### Step 1: Simplify the Integral
We start with:
\[ I = \int \frac{x^2}{1 - x^4} \, dx \]
To simplify, we can multiply the numerator and the denominator by 2:
\[ I = \frac{1}{2} \int \frac{2x^2}{1 - x^4} \, dx \]

### Step 2: Rewrite the Numerator
Next, we can express \( 2x^2 \) as:
\[ 2x^2 = (1 + x^2) - (1 - x^2) \]
Thus, we can rewrite the integral:
\[ I = \frac{1}{2} \int \frac{(1 + x^2) - (1 - x^2)}{1 - x^4} \, dx \]
This separates into two integrals:
\[ I = \frac{1}{2} \left( \int \frac{1 + x^2}{1 - x^4} \, dx - \int \frac{1 - x^2}{1 - x^4} \, dx \right) \]

### Step 3: Factor the Denominator
We can factor \( 1 - x^4 \) as:
\[ 1 - x^4 = (1 - x^2)(1 + x^2) \]
Now we can rewrite the integrals:
\[ I = \frac{1}{2} \left( \int \frac{1 + x^2}{(1 - x^2)(1 + x^2)} \, dx - \int \frac{1 - x^2}{(1 - x^2)(1 + x^2)} \, dx \right) \]
This simplifies to:
\[ I = \frac{1}{2} \left( \int \frac{1}{1 - x^2} \, dx - \int \frac{1}{1 + x^2} \, dx \right) \]

### Step 4: Integrate Each Term
Now we can integrate each term:
1. The integral \( \int \frac{1}{1 - x^2} \, dx \) can be solved using partial fractions:
\[ \int \frac{1}{1 - x^2} \, dx = \frac{1}{2} \ln \left| \frac{1 + x}{1 - x} \right| + C_1 \]
2. The integral \( \int \frac{1}{1 + x^2} \, dx \) is:
\[ \int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x) + C_2 \]

### Step 5: Combine the Results
Putting it all together, we have:
\[ I = \frac{1}{2} \left( \frac{1}{2} \ln \left| \frac{1 + x}{1 - x} \right| - \tan^{-1}(x) \right) + C \]
Thus, simplifying gives:
\[ I = \frac{1}{4} \ln \left| \frac{1 + x}{1 - x} \right| - \frac{1}{2} \tan^{-1}(x) + C \]

### Final Answer
Therefore, the final result is:
\[ I = \frac{1}{4} \ln \left| \frac{1 + x}{1 - x} \right| - \frac{1}{2} \tan^{-1}(x) + C \]