Find the indefinite integral of the function:
Start with partial fractions.
The denominator factors to: .
Thus you have .
Eliminating the denominators yields .
Multiplying this out gives .
Set , , and .
This makes , , and .
Thus you integral becomes . (I completed the square in the denominator of the second integral.)
The first integral clearly equals .
From now on, I'll be dealing only with the second integral.
Rewrite it like this: .
Now, break it up into two integrals: .
In the first integral, let . Thus . Now that integral becomes .
On the second integral, factor out the and complete the square in the denominator to get .
Then, factor a out of the denominator to get: .
Rewrite the denominator: .
Let and .
Thus your integral becomes: .
Factor out the and you have .
Adding all the parts together gives you: .