"Let $\alpha \in (0,1)$ and $\beta = {\log _e}(1 - \alpha )$. Let $${P_n}(x) = x + {{{x^2}} \over 2} + {{{x^3}} \over 3}\, + \,...\, + \,{{{x^n}} \over n},x \in (0,1)$$. Then the integral $\int\limits_0^\alpha {{{{t^{50}}} \over {1 - t}}dt}$ is equal to"

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Accepted Answer

"$\int_{0}^{\alpha} \frac{t^{50}}{1-t} d t$

$$ \begin{aligned} & = \int_{0}^{\alpha} \frac{t^{50}-1+1}{1-t}\\\\ & =-\int_{0}^{\alpha}\left(\frac{1-t^{50}}{1-t}-\frac{1}{1-t}\right) d t\\\\ & =-\int_{0}^{\alpha}\left(1+t+\ldots . .+t^{49}\right)+\int_{0}^{\alpha} \frac{1}{1-t} d t \end{aligned} $$

$=-\left(\frac{\alpha^{50}}{50}+\frac{\alpha^{49}}{49}+\ldots . .+\frac{\alpha^{1}}{1}\right)+\left(\frac{\ln (1-\mathrm{f})}{-1}\right)_{0}^{\alpha}$

$=-\mathrm{P}_{50}(\alpha)-\ln (1-\alpha)$

$=-\mathrm{P}_{50}(\alpha)-\beta$"

Let $\alpha \in (0,1)$ and $\beta = {\log _e}(1 - \alpha )$. Let $${P_n}(x) = x + {{{x^2}} \over 2} + {{{x^3}} \over 3}\, + \,...\, + \,{{{x^n}} \over n},x \in (0,1)$$. Then the integral $\int\limits_0^\alpha {{{{t^{50}}} \over {1 - t}}dt}$ is equal to

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Let $\alpha \in (0,1)$ and $\beta = {\log _e}(1 - \alpha )$. Let $${P_n}(x) = x + {{{x^2}} \over 2} + {{{x^3}} \over 3}\, + \,...\, + \,{{{x^n}} \over n},x \in (0,1)$$. Then the integral $\int\limits_0^\alpha {{{{t^{50}}} \over {1 - t}}dt}$ is equal to

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