Evaluate the following integration
`int(x^(4))/(1+x^(2))dx`

Answer

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

### Step 1: Rewrite the integrand
We start by rewriting \( x^4 \) in terms of \( x^2 \):
\[ x^4 = (x^2)^2 \]
Thus, we can express the integral as:
\[ I = \int \frac{x^4}{1+x^2} \, dx = \int \frac{x^2 \cdot x^2}{1+x^2} \, dx \]

### Step 2: Simplify the integrand
To simplify the integrand, we can manipulate the fraction:
\[ \frac{x^4}{1+x^2} = \frac{x^2(1+x^2 - 1)}{1+x^2} = \frac{x^2(1+x^2) - x^2}{1+x^2} = x^2 - \frac{x^2}{1+x^2} \]
So, we can rewrite the integral as:
\[ I = \int \left( x^2 - \frac{x^2}{1+x^2} \right) \, dx \]

### Step 3: Split the integral
Now, we can split the integral into two parts:
\[ I = \int x^2 \, dx - \int \frac{x^2}{1+x^2} \, dx \]

### Step 4: Evaluate the first integral
Using the power rule for integration, we find:
\[ \int x^2 \, dx = \frac{x^3}{3} \]

### Step 5: Evaluate the second integral
For the second integral, we can simplify \( \frac{x^2}{1+x^2} \):
\[ \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} \]
Thus, we can rewrite the integral:
\[ \int \frac{x^2}{1+x^2} \, dx = \int \left( 1 - \frac{1}{1+x^2} \right) \, dx = \int 1 \, dx - \int \frac{1}{1+x^2} \, dx \]
Calculating these integrals gives:
\[ \int 1 \, dx = x \quad \text{and} \quad \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \]
So, we have:
\[ \int \frac{x^2}{1+x^2} \, dx = x - \tan^{-1}(x) \]

### Step 6: Combine results
Now we can combine the results of our integrals:
\[ I = \frac{x^3}{3} - \left( x - \tan^{-1}(x) \right) \]
This simplifies to:
\[ I = \frac{x^3}{3} - x + \tan^{-1}(x) \]

### Final Result
Thus, the final result for the integral is:
\[ I = \frac{x^3}{3} - x + \tan^{-1}(x) + C \]
where \( C \) is the constant of integration.