Question Number 29552 by abdo imad last updated on 09/Feb/18
$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\sqrt{{x}+\mathrm{1}}\:−\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)}{dx}\:. \\ $$
Commented by prof Abdo imad last updated on 11/Feb/18
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\sqrt{{x}+\mathrm{1}}\:\:−\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)}\:{the}\:{ch}.\sqrt{{x}+\mathrm{1}}\:={t}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{\infty} \:\:\frac{{t}−\mathrm{1}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right){t}^{\mathrm{2}} }\mathrm{2}{tdt}\:=\:\int_{\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{2}}{{t}\left({t}+\mathrm{1}\right)}{dt} \\ $$$$=\:\mathrm{2}\int_{\mathrm{1}} ^{\infty} \:\:\left(\:\frac{\mathrm{1}}{{t}}\:−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt}\:=\mathrm{2}\left[{ln}\mid\frac{{t}}{{t}+\mathrm{1}}\mid_{\mathrm{1}} ^{\infty} \:\right. \\ $$$${I}=\:\mathrm{2}\left(\mathrm{0}−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)=\mathrm{2}{ln}\mathrm{2}. \\ $$
Answered by ajfour last updated on 10/Feb/18
![let t^2 =x+1 2tdt = dx I = ∫_1 ^( ∞) (((t−1)2tdt)/((t^2 −1)t^2 )) =2∫_1 ^( ∞) (dt/(t(t+1))) =2∫_1 ^( ∞) (dt/((t+(1/2))^2 −((1/2))^2 )) =2ln ((t/(t+1)))∣_0 ^∞ =2lim_(t→∞) [ln (1−(1/(t+1)))]+2ln 2 = 2ln 2 .](https://www.tinkutara.com/question/Q29606.png)
$${let}\:\:\:\:{t}^{\mathrm{2}} ={x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}{tdt}\:=\:{dx} \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{\:\:\infty} \:\frac{\left({t}−\mathrm{1}\right)\mathrm{2}{tdt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right){t}^{\mathrm{2}} }\:=\mathrm{2}\int_{\mathrm{1}} ^{\:\:\infty} \frac{{dt}}{{t}\left({t}+\mathrm{1}\right)} \\ $$$$\:\:=\mathrm{2}\int_{\mathrm{1}} ^{\:\:\infty} \frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\:\:=\mathrm{2ln}\:\left(\frac{{t}}{{t}+\mathrm{1}}\right)\mid_{\mathrm{0}} ^{\infty} \:\:=\mathrm{2}\underset{{t}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{ln}\:\left(\mathrm{1}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right)\right]+\mathrm{2ln}\:\mathrm{2} \\ $$$$\:\:=\:\mathrm{2ln}\:\mathrm{2}\:. \\ $$