3 Tough improper integrals Some weird cases of improper integrals and probably the worse way to evaluate them.
(sorry im posting 3 integrals even though they are pretty tough, but they are all very related and half my problem is how to layout my solution)

I. 

$$\int^{\infty}_{0} \frac {x^{1/2}}{e^{x}-1}dx.$$ 
This one I had to go all cheese no skill. First clearly at infinity this function is dominated by ${e^{-x}}$, whereas it tends to $0$ as $x \to \infty$. By the comparison test, the integral converges at $\infty$; the real problem happens at 0 as we have $\frac{0}{0}$. To tackle this issue, I used a Taylor expansion here to the order, thus $e^{x} \approx 1+x$ which leads us to $\frac {x^{1/2}}{1-x-1}=\frac {1}{x^{1/2}}$ as $x \to 0$ which is a convergent behavior. Thus $\int^{\infty}_{0} \frac {x^{(1/2)}}{e^{x}-1}$  converges.

II. 

$$\int^{\infty}_{0} \frac{\sin(\frac{1}{x})}{x^{1/5}}dx$$ 
As $x \to 0$, since $\sin ({1}/{x})$ always lies in the interval $[-1,1]$, the integrand is clearly dominated by $1/x^{1/5}$ and thus converges as $x$ goes to $0$. (This part was super sketchy to me because the function gets there in a funny little hop kind of way)
More interesting is how to tackle the limit as $x \to \infty$. Here I did a change of variables: let $u=\frac {1}{x}$ thus $du=-1/x^{2}$ and we have $-1/x^2$ and $1/x^{1/5}$ which leads to $u^{11/5}\sin(u)$ since, as $x$ goes to infinity, $1/x \to 0$. 
Now, in the limit $\lim_{u \to 0} u^{11/5} \sin(u)$, clearly $\sin(u)$ is dominated by $u^{11/5}$. Thus we have $\lim_{u \to 0} u^{11/5}=0$ thus the limit converges on both ends.
Biggest problem with above is that it seems like it shouldn't work; but it also has the problem that I have no idea what rules I used to show this (assuming that this is even a plausible answer).

III.

$$\int^{\infty}_{-\infty} \frac {e^{x}}{e^{x}+x^{2}}dx$$
This one has given me a lot of trouble. A taylor series expansion can't be used for obvious reasons at $\infty$; I tried a funny method of factoring but that didn't yield much that made sense. Lastly, I tried dividing by $e^{x}$.
This yields $\int^{\infty}_{-\infty} \frac {1}{e^{x}+x^{2}}dx$ which clearly has the limit as $x \to \infty=0$ and the limit $x \to -\infty=0$ but $e^{x}$ as $x \to \infty$ clearly diverges.
I have tried computing directly $\lim_{x \to \infty} \frac {e^{x}}{e^{x}+x^{2}} =1$ but the other side doesn't lend itself well to this aproach.
Basically what I want is criticism on the three proofs, on how to make them acceptable as a correct answer and also on how to prove that #3 diverges (as I am fairly certain it does, I just can't pin down why) thanks for the help.
 A: It is usually good to break up an integral into parts, so that each part has at most one "bad" feature.
For your third integral, break up into the integrals from $-\infty$ to $0$, and from $0$ to $\infty$.
Let's look at $\displaystyle \int_0^\infty \frac{e^x}{e^x+x^2}\,dx$. Informally, for large $x$, the $x^2$ term is utterly negligible in comparison with $e^x$. So for big $x$ our function is nearly $1$, and of course the integral will blow up.
To be more formal, divide top and bottom of our integrand by $e^x$. We get 
$$\dfrac{1}{1+\frac{x^2}{e^x}}.$$ 
Since $\lim_{x\to\infty} \frac{x^2}{e^x}=0$, for large enough $x$ we have $\frac{x^2}{e^x}\lt \frac{1}{2}$. So for large enough $x$, the integrand is $\gt \frac{2}{3}$.
The integral $\int_0^\infty \frac{2}{3}\,dx$ clearly diverges, so by Comparison so does $\displaystyle\int_0^\infty \frac{e^x}{e^x+x^2}\,dx$.
For your second integral, break up as you did. The ideas ou used are fine. 
We will prove that both integrals converge absolutely, meaning that if we take absolute values we still have convergence. By a standard theorem, absolute convergence implies convergence.
For the integral from $0$ to $1$, note that $|\sin(1/x)|\le 1$. Thus 
$$\left|\frac{\sin(1/x)}{x^{1/5}}\right|\le \frac{1}{x^{1/5}}.$$
By a standard theorem, $\int_0^1\frac{dx}{x^{1/5}}$ converges, so by Comparison our integral from $0$ to $1$ convrges absolutely, and hence converges.
For the integral from $1$ to $\infty$, I would suggest using the inequality $|\sin t|\le |t|$.  For $x\ge 1$, this implies that $0\lt \sin(\pi/x)\lt \frac{\pi}{x}$.
Thus in our interval, our function is between $0$ and $\frac{\pi}{x^{6/5}}$. The rest is done by Comparison. 
For the first integral, again the intuition is good.  There could be improvement in the detail.
Break up as the integral from $0$ to $1$, plus the integral from $1$ to\infty$.
For the integral from $1$ to $\infty$, true, the $e^x$ term at the bottom will cush the top. Maybe you could use the Limit Comparison Test, comparing with $\frac{1}{e^{x/2}}$.
Or else observe that for large enough $x$ (and it happens pretty early) the bottom is bigger than $(1/2)e^{x}$. for large $x$, the top is $\lt e^{x/2}$. So the ratio, for large $x$, is $\lt \frac{2}{e^{x/2}}$.
For the part from $0$ to $1$, use the fact that $\lim_{x\to 0}\frac{e^x-1}{x}=1$, or more or less equivalently use the MacLaurin series for $e^x$ to get an idea of the size of $e^x-1$. 
For $x$ close enough to $0$, $x/2\le e^x-1$. so our integrand is $\le \frac{2}{x^{1/2}}$, and by Comparison with a standard integral we have convergence. 
