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Question Number 33677 by math khazana by abdo last updated on 21/Apr/18

calculate ∫_0 ^1 ((xlnx)/(x−1))dx .

calculate∫01xlnxx−1dx.

Commented by math khazana by abdo last updated on 30/Apr/18

I =∫_0 ^1 (((x−1+1))/(x−1))ln(x)dx = ∫_0 ^1 ln(x)x +∫_0 ^1 ((lnx)/(x−1))dx ∫_0 ^1 ln(x)dx=[xlnx −x]_(x→0) ^1 =−1 ∫_0 ^1 ((ln(x))/(x−1))dx=−∫_0 ^1 (Σ_(n=0) ^∞ x^n )ln(x)dx =−Σ_(n=0) ^∞ ∫_0 ^1 x^n ln(x)dx by parts A_n = ∫_0 ^1 x^n ln(x)dx=[(1/(n+1))x^(n+1) ln(x)]_0 ^1 −∫_0 ^1 (1/(n+1)) x^n dx =−(1/(n+1)) ∫_0 ^1 x^n dx=−(1/((n+1)^2 )) ⇒ ∫_0 ^1 ((xln(x))/(x−1))dx= −1+Σ_(n=0) ^∞ (1/((n+1)^2 )) =Σ_(n=1) ^∞ (1/n^2 )−1 ∫_0 ^1 ((xln(x))/(x−1))dx = (π^2 /6) −1 .

I=∫01(x−1+1)x−1ln(x)dx=∫01ln(x)x+∫01lnxx−1dx∫01ln(x)dx=[xlnx−x]x→01=−1∫01ln(x)x−1dx=−∫01(∑n=0∞xn)ln(x)dx=−∑n=0∞∫01xnln(x)dxbypartsAn=∫01xnln(x)dx=[1n+1xn+1ln(x)]01−∫011n+1xndx=−1n+1∫01xndx=−1(n+1)2⇒∫01xln(x)x−1dx=−1+∑n=0∞1(n+1)2=∑n=1∞1n2−1∫01xln(x)x−1dx=π26−1.

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