Question:

Step 1: Express the integrand as a sum of a polynomial and a simpler fraction. We begin by manipulating the integrand $\dfrac{t^{50}}{1-t}$ using polynomial division or an algebraic identity. We know that for any positive integer $n$, the sum of a geometric series is $1+t+t^2+\dots+t^{n-1} = \dfrac{1-t^n}{1-t}$. Using this, we can rewrite the term $\dfrac{t^{50}}{1-t}$ as: $$ \frac{t^{50}}{1-t} = \frac{t^{50}-1+1}{1-t} = \frac{-(1-t^{50})}{1-t} + \frac{1}{1-t} $$ $$ \frac{t^{50}}{1-t} = -\left(\frac{1-t^{50}}{1-t}\right) + \frac{1}{1-t} $$ Applying the geometric series formula for $n=50$: $$ \frac{t^{50}}{1-t} = -(1+t+t^2+\dots+t^{49}) + \frac{1}{1-t} $$ Step 2: Substitute the transformed integrand back into the integral. Now, we substitute this expression back into the given integral: $$ \int_0^\alpha\dfrac{t^{50}}{1-t}\,dt = \int_0^\alpha \left( -(1+t+t^2+\dots+t^{49}) + \frac{1}{1-t} \right) \,dt $$ Step 3: Split the integral and evaluate each part separately. We can split the integral into two parts due to linearity: $$ \int_0^\alpha\dfrac{t^{50}}{1-t}\,dt = -\int_0^\alpha (1+t+t^2+\dots+t^{49}) \,dt + \int_0^\alpha \frac{1}{1-t} \,dt $$ First, evaluate the integral of the polynomial term: $$ -\int_0^\alpha (1+t+t^2+\dots+t^{49}) \,dt = -\left[ t + \frac{t^2}{2} + \frac{t^3}{3} + \dots + \frac{t^{50}}{50} \right]_0^\alpha $$ $$ = -\left( \left(\alpha + \frac{\alpha^2}{2} + \frac{\alpha^3}{3} + \dots + \frac{\alpha^{50}}{50}\right) - (0) \right) $$ By the definition given in the problem, $P_n(x)=x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\cdots+\dfrac{x^n}{n}$, so this part is equal to $-P_{50}(\alpha)$. Next, evaluate the integral of the fractional term: $$ \int_0^\alpha \frac{1}{1-t} \,dt = \left[ -\log_e|1-t| \right]_0^\alpha $$ $$ = (-\log_e|1-\alpha|) - (-\log_e|1-0|) $$ Since $\alpha \in (0,1)$, $1-\alpha$ is positive, so $|1-\alpha|=1-\alpha$. Also, $\log_e(1)=0$. $$ = -\log_e(1-\alpha) - (-\log_e(1)) = -\log_e(1-\alpha) - 0 = -\log_e(1-\alpha) $$ The problem defines $\beta=\log_e(1-\alpha)$, so this part is equal to $-\beta$. Step 4: Combine the results from the two parts of the integral. Adding the results of the two parts, we get: $$ \int_0^\alpha\dfrac{t^{50}}{1-t}\,dt = -P_{50}(\alpha) + (-\beta) = -P_{50}(\alpha) - \beta $$ This can be written as: $$ -(\beta + P_{50}(\alpha)) $$ Step 5: Match the result with the given options. The calculated value of the integral is $-(\beta+P_{50}(\alpha))$. This matches Option 2. The final answer is $\boxed{-(\beta+P_{50}(\alpha))}$.