I do not understand improper integrals. Is $$ \int_1^e \frac{ \mathrm{dx}}{x(\ln x)^{1/2}}$$ an improper integral? Is $$ \int_0^2 \frac{\mathrm{dx}}{x^2+6x+8}$$ an improper integral?
For both I need to evaluate the integral or show that it's divergent. I have no idea what to do and need serious help for math homework since we never covered this in our Calc 1 class.
On
Improper integrals occur in primarily two ways: an bound that goes off to infinity or a bound where the function goes off to infinity (infinitely wide vs. infinitely tall). Clearly anything of the form$$\int_0^\infty f(x)\mathrm{d}x$$ is improper and we have to really consider what happens as we approach infinity. Which should scream limit! So we set it up as follows:
$$\lim_{b\to\infty}\int_0^b f(x)\mathrm{d}x$$ and perform the integration as normal, and then take the limit.
The other possibility is when the function in the integral does funky things on our bounds (such as going off to infinity or dividing by zero, etc.) One example is this simple integral:
$$\int_0^1 \frac{\mathrm{d}x}{x-1}$$ This could potentially have an infinity area (look at the graph) because as we get close to $x=1$ we are getting close to dividing by $0$. So we pull the same trick and we look at the integral $$\int_0^{0.9} \frac{\mathrm{d}x}{x-1}$$ and then $$\int_0^{0.99} \frac{\mathrm{d}x}{x-1}$$ and it doesn't take long to realize you're taking a limit. $$\lim_{b\to1}\int_0^b \frac{\mathrm{d}x}{x-1}$$
where this time $b$ approaches a finite value.
Now, to visit your examples: neither of them have an infinite bound and so our first type of improper integral can be ruled out. Now we see if we divide by $0$ or take the $\log$ of $0$ or other math no-no's. (As Andre corrected me) The first one has the $\ln{1}=0$ in a denominator which is the same as trying to divide by $0$ and therefore, we have an improper integral. And we would want to solve:
$$\lim_{b\to 1^+}\int_b^e \frac{\mathrm{d}x}{x\left(\ln{x}\right)^{1/2}}$$ and solve the integral first (in terms of a new variable $b$) and then take the limit, as if it had nothing to do with an integral.
In the current state, it seems that the second integral has no such "funk" and therefore is not improper.
I hope this helps clear up what an improper integral is!
EDIT: I do want to point out that if the bounds on the second integral were this:
$$\int_{-2}^0\frac{\mathrm{d}x}{x^2+6x+8}$$
then we would have an improper integral because if we look at $x=-2$ we end up with $1/0$ and that's a no-no. So we, instead, consider:
$$\lim_{b\to -2^+}\int_b^0\frac{\mathrm{d}x}{x^2+6x+8}$$
and solve the integral first (in terms of a new variable $b$) and then take the limit, as if it had nothing to do with an integral.
There are two types of improper integrals: In one type, at least one limit of integration is infinite ($\infty$ or $-\infty$); in the other type, the integrand has an infinite limit somewhere on the interval of integration. Your first integral is improper since the integrand 'blows up' as $x$ approaches 1 from the right. Your second integral is proper since the denominator is always positive (and never 0) on the interval of integration.
You should be able to do the first integral with a $u$-substitution. The second integral can be done with a partial fraction decomposition.