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Find the area bounded by the $x$ axis and the curve $$y = \frac{(x^2−x−2)}{(x^3+8)}$$ between its points of intersection with the $x$ axis.

So point of intersection are $(-1,2)$ since $$f(y)=0$$ when $x^2-x-2=0$. Not we take integral $$\int^2_{-1} \frac{(x^2−x−2)}{(x^3+8)} \Rightarrow \int \frac{A}{x+2} + \frac{Bx+C}{x^2-2x+4}$$ from here if we solve equation $A=\frac{1}{3} \quad B = \frac{2}{3} C = -\frac{5}{3}$ so $$\int \frac{\frac{1}{3}}{x+2}=\frac{1}{3}ln(x+2)$$

I'm stuck here. Could someone please help? Thanks in advance

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    $\begingroup$ Seems you are on the right track, the partial fraction decomposition has $A=1/3, B=2/3$, and $C=-5/3$ $\endgroup$ – SquirtleSquad Jun 8 '16 at 23:20
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Concerning the antiderivative, you properly found that $$I=\int \frac{(x^2−x−2)}{x^3+8}dx=\frac{1}{3}\int\frac{dx}{ x+2}+\frac{1}{3}\int\frac{2 x-5}{ x^2-2 x+4}dx$$ The first integral does not make any problem. Now for the second, consider $$J=\int\frac{2 x-5}{ x^2-2 x+4}dx=\int\frac{2 x-2-3}{ x^2-2 x+4}dx=\int\frac{2 x-2}{x^2-2 x+4}dx-3\int\frac{dx}{ x^2-2 x+4}$$ In the first you can recognize $\int\frac{du}u=\log(u)$ and we are left with $$K=\int\frac{dx}{ x^2-2 x+4}=\int\frac{dx}{ x^2-2 x+1+3}=\int\frac{dx}{ (x-1)^2+3}$$ So now change variable $$x-1=\sqrt 3 y\implies x=1+\sqrt 3 y\implies dx=\sqrt 3 dy$$ which makes $$K=\frac 1 {\sqrt 3}\int\frac {dy}{1+y^2}$$ which is well known.

I am sure that you can take it from here.

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  • $\begingroup$ So the answer is $$ \frac{1}{3}ln(x+2)+ \frac{1}{3}ln(x^2-2x+4)- \frac{3}{\sqrt{3}} arctan \frac{x-1}{\sqrt{3}} + C$$ right? $\endgroup$ – user296169 Jun 9 '16 at 10:27
  • $\begingroup$ @Margarita. You are very welcome ! Did you pick the trick ? $\endgroup$ – Claude Leibovici Jun 9 '16 at 10:44

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