# 022 Sample Final A, Problem 3

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Find the antiderivative: ${\displaystyle \int {\frac {6}{x^{2}-x-12}}\,dx.}$

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

Solution:

Step 1:
First, we factor: ${\displaystyle x^{2}-x-12=(x-4)(x+3)}$ .
Step 2:
Now we want to find the partial fraction expansion for ${\displaystyle {\frac {6}{(x-4)(x+3)}}}$ , which will have the form ${\displaystyle {\frac {A}{x-4}}+{\frac {B}{x+3}}}$.
To do this, we need to solve the equation  ${\displaystyle 6\,=\,A(x+3)+B(x-4)}$.
Plugging in  ${\displaystyle -3}$ for ${\displaystyle x}$, we find that  ${\displaystyle 6\,=\,-7B}$ , and thus  ${\displaystyle B=-{\frac {6}{7}}}$ .
Similarly, we can find ${\displaystyle A}$ by plugging in ${\displaystyle 4}$ for ${\displaystyle x}$. This yields ${\displaystyle 6=7A}$ , so ${\displaystyle A={\frac {6}{7}}}$ .
This completes the partial fraction expansion:
${\displaystyle {\frac {6}{x^{2}-x-12}}\,=\,{\frac {6}{7(x-4)}}-{\frac {6}{7(x+3)}}.}$
Step 3:
By the previous step, we have
${\displaystyle \int {\frac {6}{x^{2}-x-12}}dx\,=\,\int {\frac {6}{7(x-4)}}dx-\int {\frac {6}{7(x+3)}}dx.}$
Integrating by the rule in 'Foundations',
${\displaystyle \int {\frac {6}{x^{2}-x-12}}dx\,=\,{\frac {6}{7}}\ln(x-4)-{\frac {6}{7}}\ln(x+3).}$
Step 4:
Now, make sure you remember to add the  ${\displaystyle +C}$  to the integral at the end.
${\displaystyle {\frac {6}{7}}\ln(x-4)-{\frac {6}{7}}\ln(x+3)+C.}$