022 Sample Final A, Problem 3

3) What special integral do we have to use?

Foundations:
1) What does the denominator factor into? What will be the form of the decomposition?
2) How do you solve for the numerators?
Answer:
1) Since $x^{2}-x-12=(x-4)(x+3)$  , and each term has multiplicity one, the decomposition will be of the form: ${\frac {A}{x-4}}+{\frac {B}{x+3}}$
2) After writing the equality, ${\frac {6}{x^{2}-x-12}}={\frac {A}{x-4}}+{\frac {B}{x+3}}$, clear the denominators, and evaluate both sides at x = 4, -3, Each evaluation will yield the value of one of the unknowns.
3) We have to remember that $\int {\frac {c}{x-a}}dx=c\ln(x-a)$ , for any numbers c, a.