Question 229536
To integrate by partial fractions, you need to factor the denominator:
{{{int(5/(x^2-x-6), dx) = int(5/((x-3)(x+2)), dx)}}}
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/((x-3)(x+2))}}} as {{{A/(x-3) + B/(x+2)}}} 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/(x-3) + B/(x+2)}}}? 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:
{{{A/(x-3) + B/(x+2) = (A/(x-3))((x+2)/(x+2)) + (B/(x+2))((x-3)/(x-3)) = (A(x+2))/((x-3)(x+2)) + (B(x-3))/((x-3)(x+2)) = (A(x+2) + B(x-3))/((x-3)(x+2))}}}
And we want {{{(A(x+2) + B(x-3))/((x-3)(x+2))}}} to add up to {{{5/((x-3)(x+2))}}}. This will be true if {{{(A(x+2) + B(x-3)) = 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)<br>
We can solve this equation. Simplify:
{{{Ax + 2A + Bx -3B = 5}}}
Grouping the variable and constant terms and factoring:
{{{(Ax + Bx) + (2A + (-3B)) = 5}}}
{{{(A + B)x + (2A + (-3B)) = 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!
{{{int(5/((x-3)(x+2)), dx) = int(1/(x-3) + (-1)/(x+2), dx)}}}
The new integrals are relatively easy. They are ln's:
{{{int(5/((x-3)(x+2)), dx) = int(1/(x-3) + (-1)/(x+2), dx) = ln(abs(x-3)) - ln(abs(x+2)) + C}}}
Using a properties of logarithms and absolute values, we can combine the the two ln's:
{{{int(5/((x-3)(x+2)), dx) = int(1/(x-3) + (-1)/(x+2), dx) = ln(abs(x-3)) - ln(abs(x+2)) + C = ln(abs((x-3)/(x+2))) + C}}}