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1-x-lnt-1-t-2-dt-




Question Number 91220 by Ar Brandon last updated on 28/Apr/20
∫_1 ^x ((lnt)/(1+t^2 ))dt
1xlnt1+t2dt
Commented by abdomathmax last updated on 28/Apr/20
let take a try    if x>1    we do the changement t=(1/u) ⇒  I =−∫_(1/x) ^1  ((−lnu)/(1+(1/u^2 )))(−(du/u^2 )) =−∫_(1/x) ^1  ((lnu)/(1+u^2 ))du  =−∫_(1/x) ^1 lnu(Σ_(n=0) ^∞ (−1)^n  u^(2n) )du  =Σ_(n=0) ^∞ (−1)^(n+1)  ∫_(1/x) ^1  u^(2n)  ln u du =Σ_(n=0) ^∞ (−1)^(n+1)  U_n   U_n =∫_(1/x) ^1  u^(2n)  ln(u)du =_(by parts)   [(u^(2n+1) /(2n+1))lnu]_(1/x) ^1   −∫_(1/x) ^1  (u^(2n) /(2n+1))du =(1/(2n+1))((lnx)/x^(2n+1) )−(1/(2n+1))[(1/(2n+1))u^(2n+1) ]_(1/x) ^1   =((ln(x))/((2n+1)x^(2n+1) ))−(1/((2n+1)^2 ))(1−(1/x^(2n+1) )) ⇒  I =−lnx Σ_(n=0) ^∞  (((−1)^n )/((2n+1)x^(2n+1) )) +Σ_(n=0) ^∞  (((−1)^n )/((2n+1)^2 ))  −Σ_(n=0) ^∞  (((−1)^n )/((2n+1)^2 x^(2n+1) ))  rest to calculate those sums  if x<1   I =−∫_x ^1   ((lnt)/(1+t^2 ))dt  =−∫_x ^1   lnt(Σ_(n=0) ^∞ (−1)^n  t^(2n) )dt  =−Σ_(n=0) ^∞  (−1)^n  ∫_x ^1  t^(2n)  lnt dt   and we follow the   same way ....be continued...
lettakeatryifx>1wedothechangementt=1uI=1x1lnu1+1u2(duu2)=1x1lnu1+u2du=1x1lnu(n=0(1)nu2n)du=n=0(1)n+11x1u2nlnudu=n=0(1)n+1UnUn=1x1u2nln(u)du=byparts[u2n+12n+1lnu]1x11x1u2n2n+1du=12n+1lnxx2n+112n+1[12n+1u2n+1]1x1=ln(x)(2n+1)x2n+11(2n+1)2(11x2n+1)I=lnxn=0(1)n(2n+1)x2n+1+n=0(1)n(2n+1)2n=0(1)n(2n+1)2x2n+1resttocalculatethosesumsifx<1I=x1lnt1+t2dt=x1lnt(n=0(1)nt2n)dt=n=0(1)nx1t2nlntdtandwefollowthesameway.becontinued
Commented by Ar Brandon last updated on 28/Apr/20
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Commented by mathmax by abdo last updated on 29/Apr/20
thankx
thankx

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