### Quiz 8.5:

Question:

Integrate $$\int \frac{2x^3-4x^2-x-3}{x^2-2x-3}dx$$

Solution:

$$\int \frac{2x^3-4x^2-x-3}{x^2-2x-3}dx = \int 2x dx + \int \frac{A}{x+1}+\frac{B}{x-3}dx$$

$$\frac{5x-3}{(x+1)(x-3)}=\frac{A}{x+1}+\frac{B}{x-3}$$
$$5x-3 = A(x-3)+B(x+1)$$
Letting $$x=3$$, we get that $$4B=12$$ or $$B=3$$
Letting $$x=-1$$, we get that $$-4A=-8$$ or $$A=2$$

$$= \int 2x dx + \int \left ( \frac{2}{x+1}+\frac{3}{x-3} \right ) dx$$
$$= x^2 +2 \ln |x+1| +3\ln |x-3| + C$$