divergence of $\int_{2}^{\infty}\frac{dx}{x^{2}-x-2}$ i ran into this question and im sitting on it for a long time.
why does this integral diverge:
$$\int_{2}^{\infty}\frac{dx}{x^{2}-x-2}$$
thank you very much in advance.
yaron.
 A: $$\int_{2}^{\infty}\frac{dx}{x^{2}-x-2} = \int_{2}^{\infty}\frac{dx}{(x - 2)(x+1)}$$
Hence at $x = 2$, the integrand is undefined (the denominator "zeroes out" at $x = 2, x = -1$. So $x = 2$ is not in the domain of the integrand. Although we could find the indefinite integral, e.g., using partial fractions, the indefinite integral of the term with denominator $(x - 2)$ would be $$A\ln|x - 2| + C$$ which is also undefined when $x = 2.$
Recall, the limits of integration are exactly those: limits.  $$\lim_{x\to 2} (A\ln|x - 2| + C) = -\infty$$ and hence, since the limit diverges, so too the integral diverges.
A: $$\frac1{x^2-x-2}=\frac13\left(\frac1{x-2}-\frac1{x+1}\right)\implies$$
$$\int\limits_2^\infty\frac{dx}{x^2-x-2}=\left.\lim_{b\to\infty\,,\,\epsilon\to +}\frac13\log\frac{x-2}{x+1}\right|_{2+\epsilon}^b=\frac13\left[\lim_{b\to\infty}\log\frac{b-2}{b+1}-\lim_{\epsilon\to 0^+}\log\frac{\epsilon}{3+\epsilon}\right]=\infty$$
A: Notice the following:
$(x^2 - x - 2) = (x - 2)(x + 1)$
Therefore notice that:
$\frac{2}{x^2 - x - 2} = \frac{2}{(x-2)(x+1)} $
From here we can do a partial fraction decomposition which basically means we want to find:
$A, B$ such that $\frac{A}{x-2} + \frac{B}{x+1} = \frac{2}{x^2 - x - 2}$
This alternatively means that:
$A(x+1) + B(x-2) = 2$
And therefore:
$Ax + Bx = 0 ==> A + B = 0$
$A - 2B = 2$
We solve this system of 2 linear equations in 2 unknowns to find that:
$A = \frac{2}{3}, B = -\frac{2}{3}$
So finally we have:
$\frac{2}{x^2 - x - 2} = \frac{\frac{2}{3}}{x-2} - \frac{\frac{2}{3}}{x+1}  $
From here it is obvious that by integrating this expression you are integrating over the asymptote of x - 2 and therefore the sum diverges
Hope that made sense!
A: Use partial fractions and note that $$\cfrac 1{(x-2)(x+1)} = \cfrac13\left(\cfrac1{x-2}-\cfrac 1{x+1}\right)$$.
