Question Number 40128 by maxmathsup by imad last updated on 16/Jul/18
$${calculate}\:\:\:\int_{−\mathrm{2}} ^{−\mathrm{1}} \:\:\:\:\:\frac{{dt}}{{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:. \\ $$
Commented by math khazana by abdo last updated on 21/Jul/18
![I = _(t=−u) −∫_1 ^2 ((−du)/(−u(√(1+u^2 )))) =−∫_1 ^2 (du/(u(√(1+u^2 )))) changement u =sh(x) give ∫_1 ^2 (du/(u(√(1+u^2 )))) = ∫_(argsh(1)) ^(argsh(2)) ((ch(x)dx)/(sh(x)chx)) = ∫_(ln(1+(√2))) ^(ln(2+(√5))) (dx/(sh(x))) =2 ∫_(ln(1+(√2))) ^(ln(2+(√5))) (dx/(e^x −e^(−x) )) =_(e^x =t) 2 ∫_(1+(√2)) ^(2+(√5)) (1/(t−(1/t))) (dt/t) =2 ∫_(1+(√2)) ^(2+(√5)) (dt/(t^2 −1)) = ∫_(1+(√2)) ^(2+(√5)) { (1/(t−1)) −(1/(t+1))}dt =[ln∣((t−1)/(t+1))∣]_(1+(√2)) ^(2+(√5)) =ln(((1+(√5))/(3+(√5)))) −ln( ((√2)/(2+(√2)))) ⇒ I =ln(((√2)/(2+(√2))))−ln(((1+(√5))/(3+(√5)))) .](Q40438.png)
$${I}\:\:\:=\:_{{t}=−{u}} \:\:\:\:−\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\frac{−{du}}{−{u}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:=−\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{{du}}{{u}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$${changement}\:{u}\:={sh}\left({x}\right)\:{give} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\frac{{du}}{{u}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:=\:\int_{{argsh}\left(\mathrm{1}\right)} ^{{argsh}\left(\mathrm{2}\right)} \:\:\:\:\frac{{ch}\left({x}\right){dx}}{{sh}\left({x}\right){chx}} \\ $$$$=\:\int_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \:\:\frac{{dx}}{{sh}\left({x}\right)}\:=\mathrm{2}\:\int_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \:\:\:\:\frac{{dx}}{{e}^{{x}} \:−{e}^{−{x}} } \\ $$$$=_{{e}^{{x}} ={t}} \:\:\:\:\:\mathrm{2}\:\int_{\mathrm{1}+\sqrt{\mathrm{2}}} ^{\mathrm{2}+\sqrt{\mathrm{5}}} \:\:\:\:\:\frac{\mathrm{1}}{{t}−\frac{\mathrm{1}}{{t}}}\:\frac{{dt}}{{t}} \\ $$$$=\mathrm{2}\:\int_{\mathrm{1}+\sqrt{\mathrm{2}}} ^{\mathrm{2}+\sqrt{\mathrm{5}}} \:\:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}\:=\:\int_{\mathrm{1}+\sqrt{\mathrm{2}}} ^{\mathrm{2}+\sqrt{\mathrm{5}}} \:\left\{\:\frac{\mathrm{1}}{{t}−\mathrm{1}}\:−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right\}{dt} \\ $$$$=\left[{ln}\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid\right]_{\mathrm{1}+\sqrt{\mathrm{2}}} ^{\mathrm{2}+\sqrt{\mathrm{5}}} \\ $$$$={ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}+\sqrt{\mathrm{5}}}\right)\:−{ln}\left(\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)\:\:\Rightarrow \\ $$$${I}\:={ln}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)−{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}+\sqrt{\mathrm{5}}}\right)\:. \\ $$