Q: Find the non-zero constant "c" such that the following integral is convergent.
$$\int_{-1}^\infty \frac{e^{x/c}}{\sqrt{|x|}(x+2)}dx$$
Since the interval has both an infinite endpoint and discontinuity at x=0, I thought about first expressing it as a sum of improper integrals: $$\int_{-1}^0 f(x)dx + \int_0^\infty f(x)dx$$
But I'm not sure what to do next. Do I try to get it in the form $$\frac{1}{x^p}?$$
Thanks in advance.
Outline: Split the interval into $3$ parts, $-1$ to $0$, $0$ to $1$, and the rest.
For the first two integrals, note that $e^{x/c}$ is bounded above by $e^{1/c}$, and $\frac{1}{x+2}$ is bounded above by $1$. Thus for the first two integrals, the function is non-negative and bounded above by $\frac{e^{1/c}}{|x|^{1/2}}$. But we know that $\int_0^1 \frac{1}{x^{1/2}}\,dx$ converges, so by Comparison each of our first two integrals does.
Now to the third integral. If $c$ is positive, it does not converge, while if $c\lt 0$, it converges.
For the divergence when $c$ is positive, note that for any positive $a$, the function $e^{ax}$ is ultimately larger than any polynomial. In symbols, $$\frac{e^{x/c}}{\sqrt{x}(x+2)}\to\infty$$ as $x\to\infty$. This forces divergence.
For the convergence when $c\lt 0$, note that for $x\ge 1$ our integrand is positive and less than $e^{x/c}$. But we know that if $a$ is positive then $\int_1^\infty e^{-ax}\,dx$ converges.