An improper integral is of Type II if the integrand has an infinite discontinuity in the region of integration.

Example: $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}}$ and $\displaystyle\int_{-1}^1 \frac{dx}{x^2}$ are of Type II, since
$\displaystyle\lim_{x\to0^+}\frac{1}{\sqrt x}=\infty$ and $\displaystyle\lim_{x\to0}\frac{1}{x^2}=\infty$, and $0$ is contained in the intervals $[0,1]$ and $[-1,1]$.

We tackle these the same as Type I integrals: Integrate over a slightly smaller region, and then take a limit:

As with Type I integrals, we often need to use L'Hospital's rule to evaluate the resulting limit. Also, as before, the improper integral converges if the corresponding limit exists, and diverges if it doesn't.

If the discontinuity is in the middle of the region of integration, we need to break the integral into two pieces:$$\int_{-1}^1 \frac{dx}{x^2} = \int_{-1}^0 \frac{dx}{x^2} + \int_0^1 \frac{dx}{x^2},$$and then work out the limits for both pieces. The total integral only converges if both pieces converge.

The following video explains Type II improper integrals and works a number of examples.