Let $\alpha \in (0,1)\text{ and }\beta ={\log }_e(1-\alpha ).\text{ let }{P}_n\left(x\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+\dots +\frac{x^n}{n},x\in (0,1)$. Then the
integral ${\int }_0^{\alpha } \frac{t^{50}}{1-t}dt$ is equal to

${\int }_0^{\alpha } \frac{t^{50}}{1-t}dt={\int }_0^{\alpha } \left(\frac{1-\left(1-t^{50}\right)}{1-t}\right)dt={\int }_0^{\alpha } \frac{1}{1-t}dt-{\int }_0^{\alpha } \left(1+t+t^2+t^3+\dots +t^{49}\right)dt$
$\begin{array}{l}=-[\ln (1-t\mid ){]}_0^{\alpha }-{\left[t+\frac{t^2}{2}+\frac{t^3}{3}+\dots +\frac{t^{50}}{50}\right]}_0^{\alpha }=-\ln (1-\alpha )-\left(\alpha +\frac{{\alpha }^2}{2}+\frac{{\alpha }^3}{3}+\dots +\frac{{\alpha }^{50}}{50}\right)\\ =-\left(\beta +P_{50}\left(\alpha \right)\right)\end{array}$