Given that $$\frac{x^2 + 8x - 3}{x + 2} \equiv Ax + B + \frac{C}{x + 2}$$ where $x \in \mathbb{R}, x \neq -2$ find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 8 - 2020 - Paper 2

Using the result from part (a), we can express the integral as:

$\int (x + 6 - \frac{15}{x + 2}) \, dx$

We can now split the integral:

1. Compute the first part:

   $\int (x + 6) \, dx = \frac{x^2}{2} + 6x$

2. Compute the second part:

   $-\int \frac{15}{x + 2} \, dx = -15 \ln |x + 2|$

Putting it all together, we have:

$\int \frac{x^2 + 8x - 3}{x + 2} \, dx = \frac{x^2}{2} + 6x - 15 \ln |x + 2| + C$

To find the definite integral, we evaluate at limits suitable for our question. If we are integrating from a limit that corresponds to any specific number (say 0) to 2:

Substitute $x = 0$: $\frac{0^2}{2} + 6(0) - 15 \ln |0 + 2| = -15 \ln 2$

Then, taking the limit as $x$ approaches from both sides, we get the entire evaluation as:

$\int \frac{x^2 + 8x - 3}{x + 2} \, dx = \frac{x^2}{2} + 6x - 15 \ln |x + 2|$

The answer is in the form $a + b \ln 2$ where:

• a = evaluation constant derived from limits
• b = -15.