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Question Number 98444 by mathmax by abdo last updated on 14/Jun/20
calculate ∫_0 ^1 ((lnx)/((1+x)^2 ))dx
calculate∫01lnx(1+x)2dx
Answered by abdomathmax last updated on 14/Jun/20
I =∫_0 ^1 ((lnx)/((1+x)^2 ))dx by parts u^′ =(1/((1+x)^2 )) and v=lnx I =[(1−(1/(1+x)))lnx]_0 ^1 −∫_0 ^1 (1−(1/(1+x)))(dx/x) =[((xlnx)/(1+x))]_0 ^1 −∫_0 ^1 ((xdx)/(x(1+x))) =0−∫_0 ^1 (dx/(1+x)) =−[ln(1+x)]_0 ^1 =−ln(2) ⇒ ∫_0 ^1 ((lnx)/((1+x)^2 ))dx =−ln(2)
I=∫01lnx(1+x)2dxbypartsu′=1(1+x)2andv=lnxI=[(1−11+x)lnx]01−∫01(1−11+x)dxx=[xlnx1+x]01−∫01xdxx(1+x)=0−∫01dx1+x=−[ln(1+x)]01=−ln(2)⇒∫01lnx(1+x)2dx=−ln(2)

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