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

Integral Calculus: JEE Main 2023 Mathematics question

JEE Main 2023 · 31 Jan, Morning shift · Q3450

JEE Main 202331 Jan, Morning shiftMathematicsMediumIntegral Calculus369× since 2002

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

JEE Main 2023 Integral Calculus Question with Solution (31 Jan Morning shift, Q3450) | JEE Arc

JEE Main 2023 · 31 Jan, Morning shift · Q3450

JEE Main 202331 Jan, Morning shiftMathematicsMediumIntegral Calculus369× since 2002

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