Our calc teacher told us we should always separate the summands when the integrand has a singularity inside the interval of integration:
$$\int_{-1}^{3} \frac{dx}{(1-x)^3} = \int_{-1}^{1} \frac{dx}{(1-x)^3} + \int_{1}^{3} \frac{dx}{(1-x)^3}$$
We know that the antiderivative of the integrand is
$$\int \frac{dx}{(1-x)^3} = \frac{1}{2(1-x)^2} + C$$
But if we take both limits, we find that
$$\lim_{a\to 1^+}\int_{a}^{3} \frac{dx}{(1-x)^3} dx = \lim_{a\to 1^+} \frac{1}{2(1-x)^2}\bigg|_{x=a}^{x=3} = -\infty \\ \lim_{b\to 1^-}\int_{-1}^{b} \frac{dx}{(1-x)^3} dx = \lim_{b\to 1^-} \frac{1}{2(1-x)^2}\bigg|_{x=-1}^{x=b} = +\infty$$
So... does the improper integral diverge or converge? When we take limits, e.g., it's illegal to reason this way:
$$\lim_{x\to \infty} f(x) - g(x) = \lim_{x\to \infty} f(x) - \lim_{x\to \infty} g(x) = \infty - \infty = \text{"something"}$$
Are improper integrals any different?
Your initial reasoning is correct; this integral does not converge.
There is a way to assign a value, called the Cauchy principal value, to some integrals like this which do not converge in the traditional sense. The integral at hand has Cauchy principal value 0.