Let alpha in (0,1) and beta = log_{e}(1-alpha | vedclass

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(B) We want to evaluate the integral $I = \int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$. Using the algebraic identity $\frac{t^{50}}{1-t} = \frac{t^{50}-1+

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$-\left(\beta + P_{50}(\alpha)\right)$

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(B) We want to evaluate the integral $I = \int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$. Using the algebraic identity $\frac{t^{50}}{1-t} = \frac{t^{50}-1+1}{1-t} = \frac{-(1-t^{50})}{1-t} + \frac{1}{1-t} = -(1 + t + t^2 + \dots + t^{49}) + \frac{1}{1-t}$. Integrating term by term: $I = \int_{0}^{\alpha} -(1 + t + t^2 + \dots + t^{49}) dt + \int_{0}^{\alpha} \frac{1}{1-t} dt$. $I = -\left[ t + \frac{t^2}{2} + \dots + \frac{t^{50}}{50} \right]_{0}^{\alpha} + \left[ -\ln(1-t) \right]_{0}^{\alpha}$. $I = -P_{50}(\alpha) - \ln(1-\alpha)$. Given $\beta = \log_{e}(1-\alpha)$,we have $I = -P_{50}(\alpha) - \beta = -(\beta + P_{50}(\alpha))$.

Let $\alpha \in (0,1)$ and $\beta = \log_{e}(1-\alpha)$. Let $P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^n}{n}$ for $x \in (0,1)$. Then the integral $\int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$ is equal to

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