Question Number 138516 by mnjuly1970 last updated on 14/Apr/21
![...........calculus ... ... ... (I)......... 𝛗=∫_0 ^( ∞) ((xe^x )/((1+e^x )^3 ))dx =? =(1/2)∫_0 ^( ∞) x.d(((−1)/((1+e^x )^2 )) ) =(1/2)[((−x)/((1+e^x )))]_0 ^∞ +(1/2)∫_0 ^( ∞) (dx/((1+e^x )^2 )) =(1/2)∫_0 ^( ∞) (1/(1+e^x ))dx−(1/2)∫_0 ^( ∞) (e^x /((1+e^x )^2 ))dx =−(1/2)[ln(1+e^(−x) )]_0 ^∞ −(1/2)∫_0 ^( ∞) d(((−1)/(1+e^x ))) =(1/2)ln(2)−(1/2)[((−1)/(1+e^x ))]_0 ^∞ =−(1/2)+(1/2)ln(2) ...](https://www.tinkutara.com/question/Q138516.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:………..{calculus}\:…\:…\:…\:\left(\mathrm{I}\right)……… \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{xe}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{3}} }{dx}\:=? \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} {x}.{d}\left(\frac{−\mathrm{1}}{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{−{x}}{\left(\mathrm{1}+{e}^{{x}} \right)}\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{dx}}{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\mathrm{1}+{e}^{{x}} }{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{{x}} }{\left(\mathrm{1}+{e}^{{x}} \right)^{\mathrm{2}} }{dx} \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{e}^{−{x}} \right)\right]_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} {d}\left(\frac{−\mathrm{1}}{\mathrm{1}+{e}^{{x}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{−\mathrm{1}}{\mathrm{1}+{e}^{{x}} }\right]_{\mathrm{0}} ^{\infty} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:… \\ $$