I have no idea where to start with improper integrals... I'm trying to solve the integral of 1/(x^2)-25 dx from 6 to infinity and I have no clue how to begin. Also, how do I recognize if it is divergent?
To get you started.
$\displaystyle \int_6^b {\frac{{dx}}
{{x^2 - 25}}} = \frac{1}
{{10}}\ln \left( {\frac{{b - 5}}
{{b + 5}}} \right) + \frac{{\ln (11)}}
{{10}}$.
Now find the limit, $\displaystyle \lim _{b \to \infty } \left( {\frac{1}
{{10}}\ln \left( {\frac{{b - 5}}
{{b + 5}}} \right) + \frac{{\ln (11)}}
{{10}}} \right)$
Since the integrand is positive for $\displaystyle 6\le x<\infty,$ then the limit comparison test applies with $\displaystyle \int_6^\infty\frac{dx}{x^2}$ and your integral converges.
Now, to compute its value, note that $\displaystyle \frac{1}{x^{2}-25}=\frac{1}{(x+5)(x-5)}=\frac{1}{10}\cdot \frac{(x+5)-(x-5)}{(x+5)(x-5)}=\frac{1}{10}\left( \frac{1}{x-5}-\frac{1}{x+5} \right).$
First note that
$\displaystyle \begin{aligned}\frac{1}{x^2-25} & =\frac{1}{(x+5)(x-5)}\\ & =\frac{10}{10(x-5)(x+5)}\\ & =\frac{(x+5)-(x-5)}{10(x+5)(x-5)}\\ & = \frac{1}{10(x-5)}-\frac{1}{10(x+5)}\end{aligned}$.
Now note that $\displaystyle \frac{1}{10}\int\frac{1}{x-5}-\frac{1}{x+5}\,dx=\frac{1}{10}\left(\ln\left|x-5\right|-\ln\left|x+5\right|\right)=\frac{1}{10}\ln\left|\f rac{x-5}{x+5}\right|$.
Now when you evalute $\displaystyle \int_6^{\infty}\frac{\,dx}{x^2-25}\,dx$, its the same as $\displaystyle \frac{1}{10}\int_6^{\infty}\frac{1}{x-5}-\frac{1}{x+5}\,dx=\frac{1}{10}\left.\left[\ln\left|\frac{x-5}{x+5}\right|\right]\right|_6^{\infty}$.
Spoiler:
Does this make sense?
To answer the last question. If the integral is similar to $\displaystyle \int_a^{\infty}\frac1{x^p}\,dx$, and $\displaystyle p\leq 1$, then the integral is divergent. Also, the integral can be divergent if the integrand crosses a vertical asymptote.
EDIT: Plato and Kriz beat me...