Originally Posted by

**fogel1497** 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) Mr F says: Good.

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) Mr F says: Not so good. See below.

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) Mr F says: Good.

So my final answer is .5x^2 - 2ln(x^2 + 4) - 7/x + 7/4x Mr F says: Not so good.

But this is wrong, or so says my book.