Find for which $\alpha$ the integral $\int_{0}^{1} \frac{1-x^{\alpha}}{1-x}dx$ converges
For any $\alpha$ , we have $\lim_{x \to 1^-} \frac{1-x^\alpha}{1-x}= \alpha$ (derivative of $x^\alpha$ type difference quotient). Hence we always have boundedness near $1$ for any $\alpha$. It is then $x=0$ which is likely to create a problem, if any.
We will now observe the function near $x=0$ and make conclusions. Note that if $\alpha \geq 0$ then the limit exists at $0$ anyway by the quotient rule.
In fact, we have for $\alpha < 0$ : $$ \lim_{x \to 0^+} \frac{x^{-\alpha}(1-x^{\alpha})}{(1-x)} = \lim_{x \to 0^+} \frac{x^{-\alpha}-1}{1-x} = 1 \tag{*} $$
So near $0$, $\frac{1-x^{\alpha}}{1-x}$ looks like $ x^{\alpha}$, and we know that if $\alpha \ge - 1$ that integral is not converging, while for $\alpha < -1$ it is.
Now the answer. Get rid of $\alpha \ge 0$, for which continuity holds on the interval, hence boundedness and integral convergence.
For $\alpha < 0$, we first do: $$ \int_{0}^1 \frac{1-x^{\alpha}}{1-x} dx = \int_{0}^\delta \frac{1-x^{\alpha}}{1-x} dx + \int_{\delta} ^1 \frac{1-x^{\alpha}}{1-x} dx $$
Provided LHS and RHS exist (for any $0<\delta < 1$). It is clear that the second integral on the RHS exists (for all $\alpha$)due to the continuity of the integrand on the interval. It follows that the existence of the LHS integral is down to the existence of $\int_{0}^{\delta} \frac{1-x^{\alpha}}{1-x}dx$.
Now, take say $\epsilon = 0.1$. Then, there exists a $\delta > 0$ such that $0.9 < \frac{x^{-\alpha}(1-x^{\alpha})}{1-x}<1.1 $, in other words, $$ 0.9x^{\alpha} < \frac{1-x^{-\alpha}}{1-x} < 1.1x^{\alpha} $$
for $0<x<\delta$.
If $\alpha <-1$ then the desired integral is bounded between $0.9\int_{0}^\delta x^{\alpha}dx$ and $1.1\int_{0}^\delta x^{\alpha}dx$. Otherwise the integral is unbounded, as it dominates an unbounded integral.
Hence we complete the question.
The technique to keep in mind for this problem, is that of finding asymptotically, the rate of decay of the function near the points of explosion. If the decay is not fast enough, then the integral will not converge,and vice versa, so we can use results about the decay rate to conclude.
Let's check separately convergence at $0$ and $1$, by splitting the integral at $c\in(0,1)$.
If $\alpha\ge0$ there is no question about convergence at $0$. If $\alpha<0$, you can consider $$ \int_0^c \frac{1-x^\alpha}{1-x}\,dx=\int_0^c \frac{1}{1-x}\,dx-\int_0^c \frac{x^\alpha}{1-x}\,dx $$ The first integral is not a problem, so we tackle the second one with the substitution $t=1/x$ to get $$ \int_{1/c}^\infty \frac{t^{\beta-1}}{t-1}\,dt $$ where $\beta=-\alpha>0$. This is asymptotic to $t^{\beta-2}$ and we have convergence if and only if $\beta-2<-1$, hence $\alpha>-1$.
Hence our integral converges at $0$ if and only if $\alpha>-1$.
Now note that $$ \lim_{x\to1}\frac{1-x^\alpha}{1-x}=\alpha $$ so there is no real problem with convergence at $1$.