How to take $\int \frac{x^4-2x^2+2}{x^3-2x^2-x+2}dx$?

Question

The integral itself is:

$$\int \frac{x^4-2x^2+2}{x^3-2x^2-x+2}dx$$

After long division I got:

$$\int \Big(x+2+\frac{3x^2-2}{x^3-2x^2-x+2}\Big)dx$$

And after simplifying the denumerator I got:

$$x^3-2x^2-x+2 = x^2(x-2)-1(x-2) = (x^2-1)(x-2)$$

But I am not sure about $A$ and $B$ values

Should I put $A$ $B$ or $Ax$ $Bx$ ?

Answer 1

Your PFD should start like this:

$$\frac{3x^2-2}{(x-2)(x-1)(x+1)}=\frac A{x-2}+\frac B{x-1}+\frac C{x+1}$$

And I assume you can manage the rest?

Answer 2

Hint: use that your integrand can be written as $$x+2+1/6\, \left( x+1 \right) ^{-1}+10/3\, \left( x-2 \right) ^{-1}-1/2 \, \left( x-1 \right) ^{-1} $$

Answer 3

Well, by partial fraction decomposition, we have $$\frac {3x^2-2}{x^3-2x^2-x+2} = \frac {3x^2-2}{(x-2)(x-1)(x+1)} $$ $$=\frac {1}{6 (x+1)} - \frac {1}{2 (x-1)} + \frac {10}{3 (x-2)} $$

Hope you can take it from here.