$\displaystyle \int\frac{3x^2-7x+2}{2x+4}$
I really do not even know where to start. I am very rusty, I took calculus 1 three years ago, and I am now taking calculus 2.
This is not a calculus rule, it is merely algebra.
$\displaystyle \frac{3x^2 -7x + 2}{2x + 4} = \frac{3}{2} x - \frac{13}{2} + \frac{28}{2x + 4} = \frac{3}{2} x - \frac{13}{2} + \frac{14}{x + 2} $.
Read this: Polynomial Long Division
LOL! If you are taking Calc II you are more than likely not retarded. When you don't use Calculus on an almost daily basis, you quickly forget the techniques. You should really go through and refresh yourself on these techniques.
Ok, so polynomial long division, use need it to make the degree of the polynom. in the numerator smaller than the one in denominator.
$\displaystyle \int \frac{3x^2-7x+2}{2x+4} dx$
$\displaystyle 2x + 4 |\overline{3x^2 - 7x +2}$
**See here for the long division steps, can't make it work on here
Long division polynomial.doc
The numbers in red on the attachment are the numbers you use in your new integral:
Original problem:$\displaystyle \int \frac{3x^2-7x+2}{2x+4} dx$
New integral: $\displaystyle = \int \frac{3}{2}x - 6 + \frac{28}{2x+4} dx$
I was wrong about needing partial fractions for this one. You can just integrate it normally. Some long division polynomial problems you will want to combine your answers into a new rational integral with a smaller polynomial in the numerator and then use partial fractions.
*Sorry this took me so long to get to you!
Let me know if you still need help!
You remember how to take antiderivatives, right?
Here is a simple integral just to remind you:
$\displaystyle \int 2x^2 - \frac{1}{x-2} dx$
$\displaystyle = \frac{2}{3}x^3 - ln|x-2|$
So for this one:
$\displaystyle \int \frac{3}{2}x - 6 + \frac{28}{2x+4} dx$
$\displaystyle = \frac{3}{2} \cdot\frac{x^2}{2} - 6x + 28(ln|2x+4|) + C$
$\displaystyle = \frac{3x^2}{4} - 6x + 28ln|2x+4| + C$
Got it now?