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0-xp-2-x-1-dx-p-R-




Question Number 217813 by Wuji last updated on 21/Mar/25
∫_0 ^∞ [(xp(2+x)]^(−1) dx     p∈R
0[(xp(2+x)]1dxpR
Answered by mr W last updated on 22/Mar/25
=(1/p)∫_0 ^∞ (dx/(x(2+x)))  =(1/(2p))∫_0 ^∞ ((1/x)−(1/(x+2)))dx  =(1/(2p))[ln (x/(x+2))]_0 ^∞   =((ln 2)/(2p))
=1p0dxx(2+x)=12p0(1x1x+2)dx=12p[lnxx+2]0=ln22p

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