Evaluate :
`int(x^(2)-x+2)/(x^(2)+1)dx`

Answer

To evaluate the integral

\[ \int \frac{x^2 - x + 2}{x^2 + 1} \, dx, \]

we can start by simplifying the integrand.

### Step 1: Rewrite the integrand
We can rewrite the numerator \( x^2 - x + 2 \) as follows:

\[ x^2 - x + 2 = (x^2 + 1) + (1 - x). \]

This allows us to express the integral as:

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

### Step 2: Split the integral
Now we can split the integral into two parts:

\[ \int 1 \, dx + \int \frac{1 - x}{x^2 + 1} \, dx. \]

### Step 3: Evaluate the first integral
The first integral is straightforward:

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

### Step 4: Evaluate the second integral
Now we need to evaluate

\[ \int \frac{1 - x}{x^2 + 1} \, dx. \]

We can split this integral into two parts:

\[ \int \frac{1}{x^2 + 1} \, dx - \int \frac{x}{x^2 + 1} \, dx. \]

#### Step 4.1: Evaluate \(\int \frac{1}{x^2 + 1} \, dx\)
This integral is a standard result:

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

#### Step 4.2: Evaluate \(\int \frac{x}{x^2 + 1} \, dx\)
For this integral, we can use substitution. Let

\[ u = x^2 + 1 \implies du = 2x \, dx \implies \frac{du}{2} = x \, dx. \]

Thus, we have:

\[ \int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |x^2 + 1| + C. \]

### Step 5: Combine the results
Now we can combine all parts together:

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

### Final Answer
Thus, the final result of the integral is:

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