in the case of INTEGRAL OF: (x^3 + 7) / (x^2 + 4)
If you do long division of polynomials you get:
INTEGRAL OF: X + (-4x + 7) / (x^2 + 4)
I tried doing partial fractions here, but it doesn't work. The problem i run into is that i only have one term in my denominator, it cannot be broken up. Therefore i cant do any of the 4 cases.
I thought of something new:
I broke the fraction down into:
(-4x) / (x^2 + 4) and (7) / (x^2 + 4)
The second of which i can integrate by pulling the 7 out and integrating x^(-2) + 1/4
I determined that the integral of that was 7( -x^[-1] + .25x)
Then i set this aside for later, knowing I will eventually have to add it to my other integral.
Next i looked at (-4x) / (x^2 + 4)
I pulled a -2 out and got then it looked like:
-2 INTEGRAL OF: ( 2x / ((x^2) + 4)
I decided to do u substitution for u = x^2 + 4 and my du was just 2xdx making dx=du/2x
So when i plug in u i get:
-2 INTEGRAL of: (2x / u) * (1/2x)du
The 2x's cancel out and your left with -2 *INTEGRAL of: 1/u du
This is clearly -2ln(u). I plug back in u and get -2ln(x^2 + 4)
So my final answer is .5x^2 - 2ln(x^2 + 4) - 7/x + 7/4x
But this is wrong, or so says my book.