Question Number 217813 by Wuji last updated on 21/Mar/25
![∫_0 ^∞ [(xp(2+x)]^(−1) dx p∈R](https://www.tinkutara.com/question/Q217813.png)
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left[\left({xp}\left(\mathrm{2}+{x}\right)\right]^{−\mathrm{1}} {dx}\:\:\:\right. \\ $$$${p}\in\mathbb{R} \\ $$
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))](https://www.tinkutara.com/question/Q217836.png)
$$=\frac{\mathrm{1}}{{p}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}\left(\mathrm{2}+{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{p}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}+\mathrm{2}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{p}}\left[\mathrm{ln}\:\frac{{x}}{{x}+\mathrm{2}}\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}{p}} \\ $$