Question Number 90802 by john santu last updated on 26/Apr/20
$$\left.\mathrm{1}\right)\:\underset{\mathrm{ln}\:\mathrm{2}} {\overset{\infty} {\int}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}^{{x}} −\mathrm{1}}}\:{dx}\: \\ $$$$\left.\mathrm{2}\right)\:\underset{\mathrm{ln}\:\mathrm{2}} {\overset{\infty} {\int}}\:\frac{\mathrm{1}}{{e}^{{x}} −\mathrm{1}}\:{dx}\: \\ $$
Commented by john santu last updated on 26/Apr/20
![1) u = (√(e^x −1)) ⇒ dx = ((2u du)/e^x ) ∫ _1 ^∞ ((2u)/u) (1/e^x ) du = ∫_1 ^∞ 2 (du/(u^2 +1)) = tan^(−1) (u) ]_1 ^∞ = 2((π/2)−(π/4)) = (π/2) 2)∫_(ln 2) ^∞ (e^(−x) /((e^x −1)e^(−x) )) dx = ∫_(ln 2) ^∞ (e^(−x) /(1−e^(−x) )) dx = ln (1−e^(−x) )]_(ln 2 ) ^∞ = 0 − ln ((1/2)) = ln 2](https://www.tinkutara.com/question/Q90805.png)
$$\left.\mathrm{1}\right)\:{u}\:=\:\sqrt{{e}^{{x}} −\mathrm{1}}\:\Rightarrow\:{dx}\:=\:\frac{\mathrm{2}{u}\:{du}}{{e}^{{x}} } \\ $$$$\int\underset{\mathrm{1}} {\overset{\infty} {\:}}\:\frac{\mathrm{2}{u}}{{u}}\:\frac{\mathrm{1}}{{e}^{{x}} }\:{du}\:=\:\underset{\mathrm{1}} {\overset{\infty} {\int}}\mathrm{2}\:\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\left.=\:\mathrm{tan}^{−\mathrm{1}} \left({u}\right)\:\right]_{\mathrm{1}} ^{\infty} \:=\:\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\underset{\mathrm{ln}\:\mathrm{2}} {\overset{\infty} {\int}}\frac{{e}^{−{x}} }{\left({e}^{{x}} −\mathrm{1}\right){e}^{−{x}} }\:{dx}\:=\: \\ $$$$\left.\underset{\mathrm{ln}\:\mathrm{2}} {\overset{\infty} {\int}}\frac{\mathrm{e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }\:{dx}\:=\:\mathrm{ln}\:\left(\mathrm{1}−\mathrm{e}^{−{x}} \right)\underset{\mathrm{ln}\:\mathrm{2}\:} {\overset{\infty} {\right]}} \\ $$$$=\:\mathrm{0}\:−\:\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{ln}\:\mathrm{2}\: \\ $$
Commented by mathmax by abdo last updated on 26/Apr/20
![2) ∫_(ln2) ^∞ (dx/(e^x −1)) =_(e^x =t) ∫_2 ^∞ (dt/(t(t−1))) =∫_2 ^∞ ((1/(t−1))−(1/t))dt =[ln∣((t−1)/t)∣]_2 ^(+∞) =−ln((1/2))=ln(2)](https://www.tinkutara.com/question/Q90888.png)
$$\left.\mathrm{2}\right)\:\:\int_{{ln}\mathrm{2}} ^{\infty} \:\frac{{dx}}{{e}^{{x}} −\mathrm{1}}\:=_{{e}^{{x}} ={t}} \:\:\:\int_{\mathrm{2}} ^{\infty} \:\frac{{dt}}{{t}\left({t}−\mathrm{1}\right)} \\ $$$$=\int_{\mathrm{2}} ^{\infty} \left(\frac{\mathrm{1}}{{t}−\mathrm{1}}−\frac{\mathrm{1}}{{t}}\right){dt}\:=\left[{ln}\mid\frac{{t}−\mathrm{1}}{{t}}\mid\right]_{\mathrm{2}} ^{+\infty} \:=−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)={ln}\left(\mathrm{2}\right) \\ $$