Evaluate : `int(x(1+x^(2)))/(1+x^(4))dx`.

Answer

To evaluate the integral \( \int \frac{x(1+x^2)}{1+x^4} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = x^2 \). Then, the differential \( dt = 2x \, dx \), which implies \( x \, dx = \frac{dt}{2} \).

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we have:
\[ \int \frac{x(1+x^2)}{1+x^4} \, dx = \int \frac{1}{1+t^2} \cdot \frac{dt}{2} \]
This simplifies to:
\[ \frac{1}{2} \int \frac{1+t}{1+t^2} \, dt \]

### Step 3: Separate the Integral
Now, we can separate the integral:
\[ \frac{1}{2} \left( \int \frac{1}{1+t^2} \, dt + \int \frac{t}{1+t^2} \, dt \right) \]

### Step 4: Evaluate Each Integral
1. The first integral \( \int \frac{1}{1+t^2} \, dt \) is a standard integral:
\[ \int \frac{1}{1+t^2} \, dt = \tan^{-1}(t) \]

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

### Step 5: Combine the Results
Putting it all together, we have:
\[ \frac{1}{2} \left( \tan^{-1}(t) + \frac{1}{2} \ln(1+t^2) \right) + C \]

### Step 6: Substitute Back
Now, substituting back \( t = x^2 \):
\[ \frac{1}{2} \tan^{-1}(x^2) + \frac{1}{4} \ln(1+x^4) + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{x(1+x^2)}{1+x^4} \, dx = \frac{1}{2} \tan^{-1}(x^2) + \frac{1}{4} \ln(1+x^4) + C \]