Problem with Integral of Rational Function?

Question

Integral x^2/(x^2 - x - 2) I got the answer x + ln(x+1) + C

I use long division and I end up with: X - Integral (x-2)/(x^2 - x - 2) I factor x^2 - x - 2 to (x-2)(x+1) X - Integral (x-2)/((x-2)(x+1)) and I get x - integral 1/(x+1) hence my answer but my book has completly diffrent answer x - (4/3)ln(x+2) + 1/3ln(x+1) + C

I dont understand why my answer is wrong :/

Answer 1

Not sure how you got your answer without more details of your work, but what you do is long division to get the rational function equal to $x+\frac{x+2}{(x-2)(x+1)}$ ad then use partial fractions to find the second term here is $\frac43 \cdot \frac{1}{x-2} - \frac13\cdot \frac{1}{x+1}$. Then integrate each term in this form to get the answer your book shows.

Answer 2

$\dfrac{x^2}{x^2-x -2}=1+\dfrac{x+2}{x^2-x-2}=1+\dfrac{1}{2}\bigg(\dfrac{2x-1}{x^2-x-2}+\dfrac{5}{(x-2)(x+1)}\bigg)=\\1+\dfrac{1}{2}\bigg(\dfrac{2x-1}{x^2-x-2}+\dfrac{5/3}{x-2}-\dfrac{5/3}{x+1}\bigg)$