Evaluate `int (x^(2)+x+3)/(x^(2)-x-2)dx`.

Answer

To evaluate the integral \( \int \frac{x^2 + x + 3}{x^2 - x - 2} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
We can rewrite the integrand by separating the numerator into parts that resemble the denominator.

The denominator is \( x^2 - x - 2 \). We can express the numerator \( x^2 + x + 3 \) as:
\[ x^2 + x + 3 = (x^2 - x - 2) + (2x + 5) \]
Thus, we can rewrite the integral as:
\[ \int \frac{(x^2 - x - 2) + (2x + 5)}{x^2 - x - 2} \, dx \]
This separates into two integrals:
\[ \int 1 \, dx + \int \frac{2x + 5}{x^2 - x - 2} \, dx \]

### Step 2: Evaluate the first integral
The first integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 3: Evaluate the second integral
Now we focus on the second integral \( \int \frac{2x + 5}{x^2 - x - 2} \, dx \). We can use substitution for this integral.

Let \( u = x^2 - x - 2 \). Then, differentiating gives:
\[ du = (2x - 1) \, dx \quad \Rightarrow \quad dx = \frac{du}{2x - 1} \]

We can express \( 2x + 5 \) in terms of \( u \):
\[ 2x + 5 = 2x + 5 = 2x - 1 + 6 \]
Thus, the integral becomes:
\[ \int \frac{2x - 1 + 6}{u} \cdot \frac{du}{2x - 1} = \int \frac{du}{u} + 6 \int \frac{du}{(2x - 1)u} \]

### Step 4: Solve the integrals
The first part \( \int \frac{du}{u} \) gives:
\[ \ln |u| = \ln |x^2 - x - 2| \]

For the second part, we need to express \( 2x - 1 \) in terms of \( u \). We can rewrite:
\[ \int \frac{6}{u} \, du = 6 \ln |u| = 6 \ln |x^2 - x - 2| \]

### Step 5: Combine results
Combining everything, we have:
\[ \int \frac{x^2 + x + 3}{x^2 - x - 2} \, dx = x + \ln |x^2 - x - 2| + 6 \ln |x^2 - x - 2| + C \]
This simplifies to:
\[ x + 7 \ln |x^2 - x - 2| + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{x^2 + x + 3}{x^2 - x - 2} \, dx = x + 7 \ln |x^2 - x - 2| + C \]