I need to solve this: $$\int \frac{1-x^2}{x^2+x-2}$$
I tried to simplify the Denominator in this way:
$$ x^2+x-2 = (x-1)(x+2)$$
so my new integral is:
$$\int \frac{1-x^2}{(x-1)(x+2)}$$
So now I have N°(x) > D°(x) and as far as I know in this case I should try to divide the N(x) by the D(x) but I think that this is the wrong way, I have no idea about how to procede in this integral
Any hint?
Hint : $$\frac{1-x^2}{(x-1)(x+2)} = \frac{(1-x)(1+x)}{(x-1)(x+2)}=-\frac{1+x}{x+2} = -\left ( 1 - \frac 1{x+2}\right )$$
The other solutions propose canceling the common factor which is a great method for this case, but I wish to address OP concern about degrees of numerator and denominator.
As you pointed out, the degree of numerator is bigger than denominator so you can try dividing the numerator by denominator first before doing partial fractions:
$$\frac{1-x^2}{x^2+x-2}=\frac{-(x^2+x-2)+x-1}{x^2+x-2}=-1+\frac{x-1}{x^2+x-2}$$
and now you have degree of numerator smaller than degree of denominator, and you can proceed with your partial fractions approach.
$$F=\frac{1-x^2}{(x-1)(x+2)}=\dfrac{-(x-1)(x+1)}{(x-1)(x+2)}$$
If $x-1\ne0,$ $$F=-\dfrac{x+1}{x+2}=\dfrac{1-(x+2)}{x+2}=\cdots$$
For the integrand you cancel out common terms in the numerator and denominator, so you get this:
$ f(x) = \int \frac{(1-x^2)}{(x^2+x-2)}dx = \int \frac{-x -1}{x + 2}dx = log(x+2) - x $
