Consider $$ \Pi(b, s) = \int_{0}^{b} \chi(t)\ln(s-t) dt. $$ Notice that taking derivatives with respect to $s$ can help in obtaining formulae for integrals with the derivatives of $\chi$. $$ \frac{d}{ds}\Pi(b, s) = \int_{0}^{b} \chi(t)\frac{1}{s-t} dt = -\chi(b)\ln(b-t) + \int_{0}^{b}\chi^{(1)}\ln(s-t)dt, \implies$$ $$\frac{d}{ds}\Pi(b,s)=-\frac{d}{db}\Pi(b,s)+\int_{0}^{b}\chi^{(1)}(t)\ln(s-t)dt.$$ Apply $\frac{d}{ds}$ again to notice the pattern: $$ \frac{d^{2}}{ds^2}\Pi(b,s)=-\frac{d^2}{dbds}\Pi(b,s) - (\frac{d^2}{dbds}\Pi(b,s) + \frac{d^2}{db^2}\Pi(b,s))+ \int_{0}^{b}\chi^{(2)}(t)ln(s-t)dt. $$ In general: $$ \int_{0}^{b}\chi^{(n)}(t)\ln(s-t)dt = \sum_{k=0}^{n} \binom{n}{k}\frac{d^{n}}{db^{k}ds^{n-k}}\Pi(b,s). $$
Is this formula known? I was able to come up with it by myself today.