Let α∈(0,1) and β=loge⁡(1−α). let Pn(x)=x+x22+x33+…+xnn,x∈(0,1). Then theintegral ∫0α t501−tdt is equal to

∫0α t501−tdt=∫0α 1−1−t501−tdt=∫0α 11−tdt−∫0α 1+t+t2+t3+…+t49dt=−[ln⁡(1−t∣)]0α−t+t22+t33+…+t50500α=−ln⁡(1−α)−α+α22+α33+…+α5050=−β+P50(α)

Let α∈(0,1) and β=loge⁡(1−α). let Pn(x)=x+x22+x33+…+xnn,x∈(0,1). Then theintegral ∫0α t501−tdt is equal to

∫0α t501−tdt=∫0α 1−1−t501−tdt=∫0α 11−tdt−∫0α 1+t+t2+t3+…+t49dt=−[ln⁡(1−t∣)]0α−t+t22+t33+…+t50500α=−ln⁡(1−α)−α+α22+α33+…+α5050=−β+P50(α)

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Let $α \in (0, 1) and β = \log_{e} (1 - α) . let P_{n} (x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{n}}{n}, x \in (0, 1)$ . Then the
integral $\int_{0}^{α} \frac{t^{50}}{1 - t} d t$ is equal to

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$β + P_{50} (α)$

$- (β + P_{50} (α))$

$β - P_{50} (α)$

$P_{50} (α) - β$

answer is D.

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Detailed Solution

$\int_{0}^{α} \frac{t^{50}}{1 - t} d t = \int_{0}^{α} (\frac{1 - (1 - t^{50})}{1 - t}) d t = \int_{0}^{α} \frac{1}{1 - t} d t - \int_{0}^{α} (1 + t + t^{2} + t^{3} + \dots + t^{49}) d t$
$\begin{array}{l} = - [\ln (1 - t ∣)]_{0}^{α} - {[t + \frac{t^{2}}{2} + \frac{t^{3}}{3} + \dots + \frac{t^{50}}{50}]}_{0}^{α} = - \ln (1 - α) - (α + \frac{α^{2}}{2} + \frac{α^{3}}{3} + \dots + \frac{α^{50}}{50}) \\ = - (β + P_{50} (α)) \end{array}$