In many books, articles and theses dealing with perturbative QCD it is claimed that integrals of the form $\int^1_0dx\frac{\ln^n(1-x)}{1-x}$ for $n\in\mathbb{N}$ become finite when multiplying the integrand with the factor $f(x)-f(1)$ for a function $f$ integrable on $[0,1]$. See, for example in Marco Bonvini's PhD thesis Resummation of soft and hard gluon radiation in perturbative QCD (p.140).
I do understand that for $n=0$ we can write $f(x)-f(0)=(1-x)\tilde{f}(x)$, since the function $F(x):=f(x)-f(1)$ has a zero in $x=1$. Then we have $\int^1_0dx(1-x)\tilde{f}(x)\frac{1}{1-x}=\int^1_0dx\tilde{f}(x)$.
But what about $n>0$? Let $n=1$, for example. Then we have $\int^1_0dx(f(x)-f(1))\frac{\ln(1-x)}{1-x}=\int^1_0dx\tilde{f}(x)\ln(1-x)$ and the integral still diverges. Or does it?
What you call $f(x)-f(1)$, the thesis calls $g(x)-g(1)$; I'll stick with its notation. Sec. 2.1 notes an interest in defining $[f(z)]_+$ for $f(z):=\frac{\ln^n(1-z)}{1-z}$ with $n\in\{0,\,1\}$. In particular, this log-based function isn't intended as the $g$ factor in the$$\int_0^1[f(z)]_+g(z)dz=\int_0^1dzf(z)[g(z)-g(1)]$$notation of Appendix B.3. (The above vanishes for constant $g$, including the case $g=1$ that seems to interest you.) By (B.3.3),$$\left[\frac{\ln^n(1-z)}{1-z}\right]_+=\lim_{\eta\to0^+}\left[\Theta(1-\eta-z)\frac{\ln^n(1-z)}{1-z}-\delta(1-z)\int_0^{1-\eta}dz^\prime\frac{\ln^n(1-z^\prime)}{1-z^\prime}\right],$$where the limit is performed after integration over $g$. (The thesis doesn't include these primes). Write $u:=-\ln(1-z^\prime)$ so$$\begin{align}\left[\frac{\ln^n(1-z)}{1-z}\right]_+&=\lim_{\eta\to0^+}\left[\Theta(1-\eta-z)\frac{\ln^n(1-z)}{1-z}+(-1)^{n+1}\delta(1-z)\int_0^{-\ln\eta}duu^n\right]\\&=\lim_{\eta\to0^+}\left[\Theta(1-\eta-z)\frac{\ln^n(1-z)}{1-z}+\frac{\ln^{n+1}\eta}{n+1}\delta(1-z)\right].\end{align}$$Hence$$\begin{align}\int_0^1\left[\frac{\ln^n(1-z)}{1-z}\right]_+g(z)dz&=\lim_{\eta\to0^+}\left[\int_0^{1-\eta}\frac{\ln^n(1-z)}{1-z}g(z)dz+\frac{\ln^{n+1}\eta}{n+1}g(1)\right]\\&=\lim_{\eta\to0^+}\left[(-1)^n\int_0^{-\ln\eta}u^ng(1-e^{-u})du+\frac{\ln^{n+1}\eta}{n+1}g(1)\right].\end{align}$$