SOLUTION: Evaluate the following indefinite integral: (5)/(x^(2)-x-6) dx (use partial fractions)

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Question 229536: Evaluate the following indefinite integral:
(5)/(x^(2)-x-6) dx (use partial fractions)

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
To integrate by partial fractions, you need to factor the denominator:
int%285%2F%28x%5E2-x-6%29%2C+dx%29+=+int%285%2F%28%28x-3%29%28x%2B2%29%29%2C+dx%29
Next we want to rewrite the integrand as a sum of fractions where the denominators of the fractions are the factors of the current denominator. (This is a little more complicated if you have an exponent with a exponent of 2 or more. Fortunately we are not faced with this.) So we want to rewrite 5%2F%28%28x-3%29%28x%2B2%29%29 as A%2F%28x-3%29+%2B+B%2F%28x%2B2%29 We just have to figure out what values for A and B make this work. To figure this out we need to think, how would we add A%2F%28x-3%29+%2B+B%2F%28x%2B2%29? Answer: Get the denominators the same and then add. Now we have to ask: How do we get the denominators the same? Answer: Multiply the numerator and denominator of each fraction by the other fraction's denominator. In our case:

And we want %28A%28x%2B2%29+%2B+B%28x-3%29%29%2F%28%28x-3%29%28x%2B2%29%29 to add up to 5%2F%28%28x-3%29%28x%2B2%29%29. This will be true if %28A%28x%2B2%29+%2B+B%28x-3%29%29+=+5. (If you prefer formulas instead of all this logic, here is one that will work for a fraction with two factors in the denominator with no factor having an exponent of 2 or more:
(A*(B's denominator) + B*(A's denominator) = (current numerator)

We can solve this equation. Simplify:
Ax+%2B+2A+%2B+Bx+-3B+=+5
Grouping the variable and constant terms and factoring:
%28Ax+%2B+Bx%29+%2B+%282A+%2B+%28-3B%29%29+=+5
%28A+%2B+B%29x+%2B+%282A+%2B+%28-3B%29%29+=+5
Since the right side has no x term, the left side must have one either. This will be true if
A + B = 0
The right side has a constant term of 5 so the left side must have a constant term of 5:
2A + (-3B) = 5
We can now solve this system of A and B. Solving the first equation for B we get:
B = -A
Substituting this into the other equation we get:
2A + (-3(-A)) = 5
2A + 3A = 5
5A = 5
A = 1
B = -A = -(1)
Finally we have our partial fractions!

The new integrals are relatively easy. They are ln's:

Using a properties of logarithms and absolute values, we can combine the the two ln's: