An improper integral is of Type 2 if the integrand has an infinite discontinuity in the region of integration. Examples include $\displaystyle\int_0^1 \frac{dx}{\sqrt{x}}$ and $\displaystyle\int_{-1}^1 \frac{dx}{x^2}$.

We tackle these the same as Type 1 integrals: Integrate over a slightly smaller region, and then take a limit: $$\int_0^1 \frac{dx}{\sqrt{x}} = \lim_{t \to 0^+} \int_t^1 \frac{dx}{\sqrt{x}}.$$As with Type 1 integrals, you often need to use L'Hospital's rule to evaluate the resulting limit. Also as before, the improper integral converges if the limit exists, and diverges if it doesn't. If the discontinuity is in the middle of the region of integration, you 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 2 improper integrals and works a number of examples.