Question Number 28885 by abdo imad last updated on 31/Jan/18
$${find}\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{{t}\:+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:. \\ $$
Commented by abdo imad last updated on 02/Feb/18
![let put I= ∫_(−1) ^1 (dt/(t +(√(1+t^2 )))) I= ∫_(−1) ^1 (((√(1+t^2 )) −t)/(1+t^2 −t^2 ))dt= ∫_(−1) ^1 (√(1+t^2 dt)) −∫_(−1) ^1 tdt = 2∫_0 ^1 (√(1+t^2 )) dt −0= 2 ∫_0 ^1 (√(1+t^2 )) dt the ch.t=tanx give ∫_0 ^1 (√(1+t^2 )) dt= ∫_0 ^(π/4) cosx (1+tan^2 x)dx= ∫_0 ^(π/4) (dx/(cosx)) and the ch. tan((x/2))=u give ∫_0 ^(π/4) (dx/(cosx))= ∫_0 ^((√2)−1) (1/((1−u^2 )/(1+u^2 ))) ((2du)/(1+u^2 ))=∫_0 ^((√2)−1) ((2du)/(1−u^2 )) =∫_0 ^((√2)−1) ( (1/(1+u)) +(1/(1−u)))du =[ln∣((1+u)/(1−u))∣]_0 ^((√2) −1) =ln( ((√2)/(2−(√2))))=ln( ((√2)/((√2)((√2)−1))))= −ln((√2)−1). I=−2ln((√2)−1).](Q28966.png)
$${let}\:{put}\:{I}=\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}\:+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$${I}=\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:−{t}}{\mathrm{1}+{t}^{\mathrm{2}} −{t}^{\mathrm{2}} }{dt}=\:\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} {dt}}\:−\int_{−\mathrm{1}} ^{\mathrm{1}} {tdt} \\ $$$$=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:−\mathrm{0}=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:\:{the}\:{ch}.{t}={tanx}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cosx}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{dx}}{{cosx}}\:{and}\:{the}\:{ch}. \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dx}}{{cosx}}=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \left(\:\frac{\mathrm{1}}{\mathrm{1}+{u}}\:+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right){du}\:=\left[{ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}\:−\mathrm{1}} \: \\ $$$$={ln}\left(\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)={ln}\left(\:\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}\right)=\:−{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right). \\ $$$${I}=−\mathrm{2}{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right). \\ $$