Let $\alpha \in(0,1)$ and $\beta=\log _{\mathrm{e}}(1-\alpha)$. Let $P_{n}(x)=x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\ldots . .+\frac{x^{n}}{n}, x \in(0,1)$. Then the integral $\int_{0}^{\alpha} \frac{\mathrm{t}^{50}}{1-\mathrm{t}} \mathrm{dt}$ is equal to

(B) $-\left(\beta+\mathrm{P}_{50}(\alpha)\right)$

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Definite Integrals

Let

α \in (0, 1)

and

β = lo g_{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} dt

is equal to

Answer: (B)

Write

\frac{t ^{50}}{1 - t} = \frac{t ^{50} - 1}{1 - t} + \frac{1}{1 - t} = - (1 + t + \dots + t^{49}) + \frac{1}{1 - t}

. Integrating from

0

α

, the polynomial part gives

- (α + \frac{α ^{2}}{2} + \dots + \frac{α ^{50}}{50}) = - P_{50} (α)

, while

\int_{0}^{α} \frac{d t}{1 - t} = - ln (1 - α) = - β

. Adding the two parts, the integral equals

- (β + P_{50} (α))

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option