Improper integrals

Two kinds, both of which look impossible:

• A limit of $\infty$ (infinity: no end at all), like $\displaystyle\int_1^{\infty} \dfrac{6}{x^3}\,dx$. The region goes on for ever.
• A limit where the function is undefined, like $\displaystyle\int_0^4 \dfrac{1}{\sqrt{x}}\,dx$. You can't put $x = 0$ into $\dfrac{1}{\sqrt{x}}$.

Plot twist: both can still have a perfectly ordinary answer.

1. Write it as powers and integrate as normal.
2. Never put $\infty$ in as a number. Say what each term does as $x \to \infty$ instead ($x \to \infty$ reads "as $x$ gets bigger and bigger, without end"): fractions like $\dfrac{1}{x}$ and $\dfrac{3}{x^2}$ get closer and closer to 0.
3. For a limit where the function is undefined, like $x = 0$ in $\dfrac{1}{\sqrt{x}}$, put it into the integrated expression. $2\sqrt{x}$ is fine at 0, even though $\dfrac{1}{\sqrt{x}}$ isn't.
4. Top minus bottom, as usual.