Question Number 103871 by Dwaipayan Shikari last updated on 17/Jul/20

∫_0 ^1 ((x^(98) −99x+98)/(logx))dx

Commented byDwaipayan Shikari last updated on 18/Jul/20

I(a)=∫_0 ^1 ((x^a −1)/(logx))  I^′ (a)=∫_0 ^1 (∂/∂a)(((x^a −1)/(logx)))=∫_0 ^1 ((x^a logx)/(logx))=∫_0 ^1 x^a =[(x^(a+1) /(a+1))]_0 ^1 =(1/(a+1))  ∫I^( ′) (a)=∫(1/(a+1))  I(a)=log(a+1)+C=∫((x^a −1)/(logx))  So ,  ∫_0 ^1 ((x^(98) −1)/(logx))−99∫_0 ^1 ((x−1)/(logx))=log99−99log2

Commented byDwaipayan Shikari last updated on 18/Jul/20

Kindly check my answer

Answered by Ar Brandon last updated on 17/Jul/20

Let x=e^u ⇒dx=e^u du  ⇒I=∫_∞ ^0 ((e^(99u) −99e^(2u) +98e^u )/u)du          =−∫_0 ^∞ {Σ_(k=0) ^∞ ((99^k u^(k−1) )/(k!))−99Σ_(k=0) ^∞ ((2^k u^(k−1) )/(k!))+98Σ_(k=0) ^∞ (u^(k−1) /(k!))}du          =−[Σ_(k=0) ^∞ ((99^k u^k )/(k(k!)))−99Σ_(k=0) ^∞ ((2^k u^k )/(k(k!)))+98Σ_(k=0) ^∞ (u^k /(k(k!)))]_0 ^∞