Let α ∈(0,1) and β= log e(1-α). Let Pn(x)=x+(x2/2)+(x3/3)+ ldots+(xn/n), x ∈(0,1). Then the integral ∫ limits0α (t50/1-t) d t is equal to

Question

Let α ∈(0,1) and β= log e(1-α). Let Pn(x)=x+(x2/2)+(x3/3)+ ldots+(xn/n), x ∈(0,1). Then the integral ∫ limits0α (t50/1-t) d t is equal to (A) P50(α)-β (B) -(β+P50(α)) (C) β+P50(a) (D) β-P50(α)

Tardigrade Admin · Accepted Answer

∫ limits0α (t50-1+1/1-t)=-∫ limits0α(1+ t + ldots . .+ t 49)+∫ limits0α (1/1- t ) dt =-((α50/50)+(α49/49)+ ldots . .+(α1/1))+(( ln (1- f )/-1))0α =- P 50(α)- ln (1-α) =- P 50(α)-β

Q. 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 $0 \int α \frac{t ^{50}}{1 - t} d t$ is equal to

$P_{50} (α) - β$

$- (β + P_{50} (α))$

$β + P_{50} (a)$

$β - P_{50} (α)$

$0 \int α \frac{t ^{50} - 1 + 1}{1 - t} = - 0 \int α (1 + t + \dots .. + t^{49}) + 0 \int α \frac{1}{1 - t} d t$
$= - (\frac{α ^{50}}{50} + \frac{α ^{49}}{49} + \dots .. + \frac{α ^{1}}{1}) + (\frac{l n ( 1 - f )}{- 1})_{0}^{α}$
$= - P_{50} (α) - ln (1 - α)$
$= - P_{50} (α) - β$

Q. Let α∈(0,1) and β=loge​(1−α). Let Pn​(x)=x+2x2​+3x3​+…+nxn​,x∈(0,1). Then the integral 0∫α​1−tt50​dt is equal to

P50​(α)−β

−(β+P50​(α))

β+P50​(a)

β−P50​(α)