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

∫_0 ^1 ((x^(98) −99x+98)/(logx))dx

01x9899x+98logxdx

Commented by Dwaipayan 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

I(a)=01xa1logxI(a)=01a(xa1logx)=01xalogxlogx=01xa=[xa+1a+1]01=1a+1I(a)=1a+1I(a)=log(a+1)+C=xa1logxSo,01x981logx9901x1logx=log9999log2

Commented by Dwaipayan Shikari last updated on 18/Jul/20

Kindly check my answer

Kindlycheckmyanswer

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 ^∞

Letx=eudx=euduI=0e99u99e2u+98euudu=0{k=099kuk1k!99k=02kuk1k!+98k=0uk1k!}du=[k=099kukk(k!)99k=02kukk(k!)+98k=0ukk(k!)]0

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