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Question Number 118246 by bemath last updated on 16/Oct/20

∫_0 ^1 ((ln x)/(x+1)) dx =?

∫10lnxx+1dx=?

Answered by Dwaipayan Shikari last updated on 16/Oct/20

∫_0 ^1 ((logx)/(x+1))dx=[log(x)log(x+1)]_0 ^1 −∫_0 ^1 (1/x)log(x+1) =0−∫_0 ^1 (−1)^(n+1) Σ_(n=1) ^∞ (x^(n−1) /n) =−Σ_(n=1) ^∞ (−1)^(n+1) (1/n^2 ) =−(π^2 /(12))

∫01logxx+1dx=[log(x)log(x+1)]01−∫011xlog(x+1)=0−∫01(−1)n+1∑∞n=1xn−1n=−∑∞n=1(−1)n+11n2=−π212

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