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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 $Ax$ $Bx$ ?

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    http://www.wolframalpha.com/input/?i=%E2%88%AB(x%5E4%E2%88%922x%5E2%2B2)%2F(x%5E3%E2%88%922x%5E2%E2%88%92x%2B2)dx – 2017-02-23
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    As a difference of two squares, $x^2-1=(x-1)(x+1)$. – 2017-02-23
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    Where do $A$ and $B$ come into this? – 2017-02-23
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    @Shaun it is standard PFD. – 2017-02-23
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    By the way, all your previous steps look good. – 2017-02-23

3 Answers 3

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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?

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    Indeed, I missed that $(x^2-1) = (x-1)(x+1)$, so the expression has no complex roots and all values can be found easily – 2017-02-23
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    :-) well, just one small step you've missed, but it looks like you've got the rest? And of course, do ask if you need more assistance. – 2017-02-23
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Hint: use that your integrand can be written as $$x+2+1/6\, \left( x+1 \right) ^{-1}+10/3\, \left( x-2 \right) ^{-1}-1/2 \, \left( x-1 \right) ^{-1} $$

2

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.