The given expression is:

$\int \frac{2x^2 + 7x - 3}{x - 2} \, dx$

Solution Approach:

We will solve this by performing polynomial long division on $\frac{2x^2 + 7x - 3}{x - 2}$, since the degree of the numerator is higher than that of the denominator.

Step 1: Perform polynomial long division

We divide $2x^2 + 7x - 3$ by $x - 2$.

• Divide the leading terms: $\frac{2x^2}{x} = 2x$.

• Multiply $2x$ by $x - 2$, which gives $2x^2 - 4x$.

• Subtract $(2x^2 - 4x)$ from $(2x^2 + 7x - 3)$ to get:

  $(2x^2 + 7x - 3) - (2x^2 - 4x) = 11x - 3$

• Now, divide $11x$ by $x$, which gives $11$.

• Multiply $11$ by $x - 2$, which gives $11x - 22$.

• Subtract $(11x - 22)$ from $(11x - 3)$ to get:

  $(11x - 3) - (11x - 22) = 19$

So, the result of the division is:

$2x + 11 + \frac{19}{x - 2}$

Step 2: Integrate the result

Now, the integral becomes:

$\int \left(2x + 11 + \frac{19}{x - 2}\right) dx$

We can integrate this term by term:

• $\int 2x \, dx = x^2$
• $\int 11 \, dx = 11x$
• $\int \frac{19}{x - 2} \, dx = 19 \ln|x - 2|$

Thus, the solution to the integral is:

$x^2 + 11x + 19 \ln|x - 2| + C$

where $C$ is the constant of integration.

Final Answer:

$\boxed{x^2 + 11x + 19 \ln|x - 2| + C}$

Let me know if you need further clarification or additional details!

Here are 5 related questions for further exploration:

1. How would the solution change if the denominator was $x + 2$ instead of $x - 2$?
2. How would you integrate $\frac{x^2 + 7x - 3}{x - 1}$?
3. What happens if the numerator and denominator are both cubic polynomials?
4. How would you integrate a rational function where the numerator is of a lower degree than the denominator?
5. How does the natural logarithm $\ln|x - 2|$ appear in this integration?

Tip: Polynomial long division is essential when the degree of the numerator exceeds the degree of the denominator in a rational function.