Improper integrals with polynomial in the denominator - how does p-series apply? For example, 
$$\int_2^{\infty} \frac1{x^2+2x-3} dx$$
The power in the denominator is greater than 1, so shouldn't it converge? When solving with partial fractions, it appears that it diverges. However, WolframAlpha gives me a numerical answer (log5/4). Do I need to solve it with a different method? Does it converge of diverge?
 A: \begin{align}
\int_2^\infty \frac{1}{x^2+2x-3} dx & = \frac 14 \int_2^\infty \bigg(\frac{1}{x-1} - \frac{1}{x+3} \bigg) dx \\ \tag{1}
& = \frac 14 \bigg[\ln|x-1| - \ln |x+3| \bigg]_2^\infty \\ \tag{2}
& = \frac 14 \bigg[\ln \bigg|\frac{x-1}{x+3} \bigg| \bigg ]_2^\infty \\
& = \frac 14\bigg(\ln(1) - \ln\Big(\frac 15 \Big) \bigg) \\
& = -\frac 14 \ln \Big(\frac 15\Big) \\
& = \frac 14 \ln5
\end{align}
Basically, there are many subtle things going on here.
For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".
This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing
$$\lim_{x\rightarrow \infty} \ln \bigg|\frac{x-1}{x+3} \bigg|$$
We know that 
$$\lim_{x\rightarrow \infty} \bigg|\frac{x-1}{x+3} \bigg| = 1$$
Hence
$$\lim_{x\rightarrow \infty} \ln \bigg|\frac{x-1}{x+3} \bigg| = \ln(1) = 0$$
which is where the "infinities cancel out".
This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that
$$\int_a^b \big(f(x)+g(x) \big) dx = \int_a^b f(x)dx + \int_a^b g(x)dx$$
However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.
A: When you use partial fractions you may loose convergence on each single fraction.   
To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n \to \infty$. 
