022 Sample Final A, Problem 3

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

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?
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.

Solution:

Step 1:
First, we factor $x^{2}-x-12=(x-4)(x+3)$

Step 2:
Now we want to find the partial fraction expansion for ${\frac {6}{(x-4)(x+3)}}$ , which will have the form ${\frac {A}{x-4}}+{B}{x+3}$
To do this we need to solve the equation $6=A(x+3)+B(x-4)$
Plugging in -3 for x to both sides we find that $6=-7B$   and   $B=-{\frac {6}{7}}$.
Now we can find A by plugging in 4 for x to both sides. This yields $6=7A$ , so $A={\frac {6}{7}}$
Finally we have the partial fraction expansion: ${\frac {6}{x^{2}-x-12}}={\frac {6}{7(x-4)}}-{\frac {6}{7(x+3)}}$

Step 3:
Now to finish the problem we integrate each fraction to get: $\int {\frac {6}{x^{2}-x-12}}dx=\int {\frac {6}{7(x-4)}}dx-\int {\frac {6}{7(x+3)}}dx$  to get  ${\frac {6}{7}}\ln(x-4)-{\frac {6}{7}}\ln(x+3)$

Step 4:
Now make sure you remember to add the $+C$ to the integral at the end.

Final Answer:
${\frac {6}{7}}\ln(x-4)-{\frac {6}{7}}\ln(x+3)+C$

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