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

Solution

Maths sir · 4 steps

Write t⁵⁰ as (t⁵⁰-1)+1; the t⁵⁰-1 part divides cleanly by 1-t into a geometric sum.

1. Split the numerator. Write t⁵⁰/(1-t) = 1/(1-t) - (1-t⁵⁰)/(1-t). The second fraction is a finite geometric sum, (1-t⁵⁰)/(1-t) = 1 + t + t² + … + t⁴⁹.

   ∫₀^αt⁵⁰/(1-t) dt = ∫₀^αdt/(1-t) - ∫₀^α(1 + t + … + t⁴⁹)dt

2. First integral. ∫₀^αdt/(1-t) = [-ln(1-t)]₀^α = -ln(1-α) = -β.
3. Second integral. Integrating term by term gives α + α²/2 + … + α⁵⁰/50, which is exactly P₅₀(α).
4. So. The integral is -β - P₅₀(α) = -(β + P₅₀(α)), which is option (a).

The idea to keep. When you see tⁿ/(1-t), add and subtract 1 in the numerator. You get a geometric polynomial plus 1/(1-t), and ∫ dt/(1-t) carries a minus sign.

Answer from the NTA final answer key. Solution by PrepWiser.