`int(x^2dx)/(4+x^2)`

Answer

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

### Step 1: Rewrite the Integrand
We start with the integral:
\[ I = \int \frac{x^2}{4 + x^2} \, dx \]
We can rewrite the numerator \( x^2 \) by adding and subtracting \( 4 \):
\[ I = \int \frac{x^2 + 4 - 4}{4 + x^2} \, dx = \int \frac{x^2 + 4}{4 + x^2} \, dx - \int \frac{4}{4 + x^2} \, dx \]

### Step 2: Simplify the First Integral
Now, we can separate the first integral:
\[ I = \int \frac{x^2 + 4}{4 + x^2} \, dx - 4 \int \frac{1}{4 + x^2} \, dx \]
The first part simplifies as follows:
\[ \int \frac{x^2 + 4}{4 + x^2} \, dx = \int 1 \, dx = x \]
because \( \frac{x^2 + 4}{4 + x^2} = 1 \).

### Step 3: Solve the Second Integral
Now, we need to solve the second integral:
\[ -4 \int \frac{1}{4 + x^2} \, dx \]
We recognize that \( 4 \) can be written as \( 2^2 \), so we can use the formula for the integral of \( \frac{1}{x^2 + a^2} \):
\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \]
In our case, \( a = 2 \):
\[ -4 \int \frac{1}{4 + x^2} \, dx = -4 \cdot \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) = -2 \tan^{-1} \left( \frac{x}{2} \right) \]

### Step 4: Combine the Results
Putting everything together, we have:
\[ I = x - 2 \tan^{-1} \left( \frac{x}{2} \right) + C \]
where \( C \) is the constant of integration.

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