Evaluate
`int (x^(2))/(x^(2)+5)dx`

Answer

To evaluate the integral \( \int \frac{x^2}{x^2 + 5} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand:
\[ \frac{x^2}{x^2 + 5} = \frac{x^2 + 5 - 5}{x^2 + 5} = \frac{x^2 + 5}{x^2 + 5} - \frac{5}{x^2 + 5} = 1 - \frac{5}{x^2 + 5} \]
Thus, we can express the integral as:
\[ I = \int \left( 1 - \frac{5}{x^2 + 5} \right) \, dx \]

### Step 2: Split the integral
Now we can split the integral into two parts:
\[ I = \int 1 \, dx - 5 \int \frac{1}{x^2 + 5} \, dx \]

### Step 3: Integrate the first part
The integral of 1 with respect to \( x \) is:
\[ \int 1 \, dx = x \]

### Step 4: Integrate the second part
Next, we need to evaluate the integral \( \int \frac{1}{x^2 + 5} \, dx \). We can rewrite \( 5 \) as \( \sqrt{5}^2 \):
\[ \int \frac{1}{x^2 + 5} \, dx = \int \frac{1}{x^2 + (\sqrt{5})^2} \, dx \]
Using the formula \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \), where \( a = \sqrt{5} \):
\[ \int \frac{1}{x^2 + 5} \, dx = \frac{1}{\sqrt{5}} \tan^{-1} \left( \frac{x}{\sqrt{5}} \right) + C \]

### Step 5: Combine the results
Now, substituting back into our expression for \( I \):
\[ I = x - 5 \left( \frac{1}{\sqrt{5}} \tan^{-1} \left( \frac{x}{\sqrt{5}} \right) \right) + C \]
This simplifies to:
\[ I = x - \frac{5}{\sqrt{5}} \tan^{-1} \left( \frac{x}{\sqrt{5}} \right) + C \]
Since \( \frac{5}{\sqrt{5}} = \sqrt{5} \), we can write:
\[ I = x - \sqrt{5} \tan^{-1} \left( \frac{x}{\sqrt{5}} \right) + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{x^2}{x^2 + 5} \, dx = x - \sqrt{5} \tan^{-1} \left( \frac{x}{\sqrt{5}} \right) + C \]