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Question Number 217813 by Wuji last updated on 21/Mar/25

∫_0 ^∞ [(xp(2+x)]^(−1) dx     p∈R

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left[\left({xp}\left(\mathrm{2}+{x}\right)\right]^{−\mathrm{1}} {dx}\:\:\:\right. \\ $$$${p}\in\mathbb{R} \\ $$

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))

$$=\frac{\mathrm{1}}{{p}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}\left(\mathrm{2}+{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{p}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}+\mathrm{2}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{p}}\left[\mathrm{ln}\:\frac{{x}}{{x}+\mathrm{2}}\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}{p}} \\ $$

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