= Solution
On $0<x<1$, the geometric expansion gives summands $h_n(x)=x^{n+\beta-1}-x^{n-\beta}$. Their <integrals> are
$$
\int_0^1h_n(x)\,dx
=\frac1{n+\beta}-\frac1{n+1-\beta}
=\frac{1-2\beta}{n^2+n+\beta-\beta^2}.
$$
Both powers are integrable because $0<\beta<1$. If $\beta<1/2$, each $h_n$ is nonnegative, since its first exponent is smaller. Their <integrals> form a convergent nonnegative series: the $n=0$ term is finite and subsequent terms are bounded by a constant times $n^{-2}$. The proved series-integration rule applies.
If $\beta>1/2$, apply precisely that rule to $-h_n$, which are nonnegative, and negate the result. At $\beta=1/2$ all summands and the integrand are zero. Thus in every case the <integral> is absolutely integrable and
$$
\boxed{\int_0^1\frac{x^{\beta-1}-x^{-\beta}}{1-x}\,dx
=\sum_{n=0}^{\infty}\frac{1-2\beta}{n^2+n+\beta-\beta^2}.}
$$
The common sign of the summands is the reason no unproved theorem for integrating arbitrary signed series is needed.
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