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:
\[ \frac{x^2}{x^2 + 5} = \frac{x^2 + 5 - 5}{x^2 + 5} = \frac{x^2 + 5}{x^2 + 5} - \frac{5}{x^2 + 5} \]
This simplifies to:
\[ 1 - \frac{5}{x^2 + 5} \]

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

### Step 3: Evaluate the first integral
The first integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 4: Evaluate the second integral
For the second integral, we recognize that \(5\) can be written as \(\sqrt{5}^2\):
\[ \int \frac{1}{x^2 + 5} \, dx = \int \frac{1}{x^2 + (\sqrt{5})^2} \, dx \]
Using the formula for the integral of \(\frac{1}{x^2 + a^2}\):
\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \]
we have \(a = \sqrt{5}\):
\[ \int \frac{1}{x^2 + 5} \, dx = \frac{1}{\sqrt{5}} \tan^{-1} \left(\frac{x}{\sqrt{5}}\right) + C \]

### Step 5: Combine the results
Now we can combine our results:
\[ \int \frac{x^2}{x^2 + 5} \, dx = x - 5 \left(\frac{1}{\sqrt{5}} \tan^{-1} \left(\frac{x}{\sqrt{5}}\right)\right) + C \]
This simplifies to:
\[ x - \sqrt{5} \tan^{-1} \left(\frac{x}{\sqrt{5}}\right) + C \]

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