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

Answer

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

### Step 1: Rewrite the integrand
We can rewrite the integrand by adding and subtracting 4 in the numerator:

\[ \frac{x^2}{x^2 + 4} = \frac{x^2 + 4 - 4}{x^2 + 4} = \frac{x^2 + 4}{x^2 + 4} - \frac{4}{x^2 + 4} \]

This simplifies to:

\[ 1 - \frac{4}{x^2 + 4} \]

### Step 2: Set up the integral
Now we can rewrite the integral:

\[ \int \frac{x^2}{x^2 + 4} \, dx = \int \left( 1 - \frac{4}{x^2 + 4} \right) \, dx \]

### Step 3: Split the integral
We can split the integral into two parts:

\[ \int \left( 1 - \frac{4}{x^2 + 4} \right) \, dx = \int 1 \, dx - 4 \int \frac{1}{x^2 + 4} \, dx \]

### Step 4: Integrate the first part
The integral of 1 is straightforward:

\[ \int 1 \, dx = x \]

### Step 5: Integrate the second part
For the second integral, we recognize that \( x^2 + 4 \) can be rewritten as \( x^2 + 2^2 \). We can use the formula for the integral of \( \frac{1}{x^2 + a^2} \):

\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]

In our case, \( a = 2 \):

\[ \int \frac{1}{x^2 + 4} \, dx = \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) \]

### Step 6: Combine the results
Now we combine the results from both integrals:

\[ \int \frac{x^2}{x^2 + 4} \, dx = x - 4 \left( \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) \right) + C \]

This simplifies to:

\[ x - 2 \tan^{-1} \left( \frac{x}{2} \right) + C \]

### Final Answer
Thus, the final answer is:

\[ \int \frac{x^2}{x^2 + 4} \, dx = x - 2 \tan^{-1} \left( \frac{x}{2} \right) + C \]

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