Question Number 89714 by abdomathmax last updated on 18/Apr/20
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 19/Apr/20
![parametric method let f(t)=∫_0 ^∞ (dx/(t+(√(2+x^2 )))) with t>0 f^′ (t)=−∫_0 ^∞ (dx/((t+(√(2+x^2 )))^2 )) ⇒∫_0 ^∞ (dx/((t+(√(2+x^2 )))^2 )) =−f^′ (t) changement x =(√2)sh(u) give f(t)=∫_0 ^∞ (((√2)ch(u)du)/(t +(√2)ch(u))) =(√2)∫_0 ^∞ (((e^u +e^(−u) )/2)/(t+(√2)×((e^u +e^(−u) )/2)))du =(√2)∫_0 ^∞ ((e^u +e^(−u) )/(2t+(√2)e^u +(√2)e^(−u) ))du =_(e^u =z) (√2)∫_1 ^∞ ((z+z^(−1) )/(2t+(√2)z +(√2)z^(−1) ))×(dz/z) =(√2)∫_1 ^(+∞) ((z+z^(−1) )/(2tz +(√2)z^2 +(√2))) dz =(√2)∫_1 ^(+∞) ((z^2 +1)/(z((√2)z^2 +2tz +(√2))))dz decoposition of g(z)=((z^2 +1)/(z((√2)z^2 +2tz +(√2)))) (√2)z^2 +2tz +(√2)=0 →Δ^′ =t^2 −2<0 if t<(√2) z_1 =((−t+i(√(2−t^2 )))/( (√2))) and z_2 =((−t−i(√(2−t^2 )))/( (√2))) g(z)=((z^2 +1)/( (√2)z(z−z_1 )(z−z_2 ))) =(a/z) +(b/(z−z_1 )) +(c/(z−z_2 )) a =(1/( (√2)z_1 z_2 )) =(1/( (√2))) ,b =((z_1 ^2 +1)/( (√2)z_1 (√2)i(√(2−t^2 )))) =((1+z_1 ^2 )/(2iz_1 (√(2−t^2 )))) b=((z_2 ^2 +1)/( (√2)z_2 (−(√2)i(√(2−t^2 ))))) =((1+z_2 ^2 )/(−2iz_2 (√(2−t^2 )))) ⇒ ∫_1 ^(+∞) g(z)dz =∫_1 ^(+∞) ((a/z)+(b/(z−z_1 )) +(c/(z−z_2 )))dz =[aln∣z∣ +bln∣z−z_1 ∣ +cln∣z−z_2 ∣]_1 ^(+∞) rest to simplify b and c be continued...](https://www.tinkutara.com/question/Q89825.png)
$${parametric}\:{method}\:{let}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{t}+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}\:\:{with}\:{t}>\mathrm{0} \\ $$$${f}^{'} \left({t}\right)=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({t}+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({t}+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:=−{f}^{'} \left({t}\right) \\ $$$$\:{changement}\:{x}\:=\sqrt{\mathrm{2}}{sh}\left({u}\right)\:{give} \\ $$$${f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\sqrt{\mathrm{2}}{ch}\left({u}\right){du}}{{t}\:+\sqrt{\mathrm{2}}{ch}\left({u}\right)}\:=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\frac{{e}^{{u}} +{e}^{−{u}} }{\mathrm{2}}}{{t}+\sqrt{\mathrm{2}}×\frac{{e}^{{u}} +{e}^{−{u}} }{\mathrm{2}}}{du} \\ $$$$=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{{u}} +{e}^{−{u}} }{\mathrm{2}{t}+\sqrt{\mathrm{2}}{e}^{{u}} \:+\sqrt{\mathrm{2}}{e}^{−{u}} }{du}\:=_{{e}^{{u}} ={z}} \:\:\sqrt{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \:\:\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}{t}+\sqrt{\mathrm{2}}{z}\:+\sqrt{\mathrm{2}}{z}^{−\mathrm{1}} }×\frac{{dz}}{{z}} \\ $$$$=\sqrt{\mathrm{2}}\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}{tz}\:+\sqrt{\mathrm{2}}{z}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}}\:{dz} \\ $$$$=\sqrt{\mathrm{2}}\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{{z}\left(\sqrt{\mathrm{2}}{z}^{\mathrm{2}} \:+\mathrm{2}{tz}\:+\sqrt{\mathrm{2}}\right)}{dz}\:{decoposition}\:{of}\:{g}\left({z}\right)=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{{z}\left(\sqrt{\mathrm{2}}{z}^{\mathrm{2}} \:+\mathrm{2}{tz}\:+\sqrt{\mathrm{2}}\right)} \\ $$$$\sqrt{\mathrm{2}}{z}^{\mathrm{2}} \:+\mathrm{2}{tz}\:+\sqrt{\mathrm{2}}=\mathrm{0}\:\rightarrow\Delta^{'} ={t}^{\mathrm{2}} −\mathrm{2}<\mathrm{0}\:\:{if}\:{t}<\sqrt{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\frac{−{t}+{i}\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}}}\:\:{and}\:{z}_{\mathrm{2}} =\frac{−{t}−{i}\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}}} \\ $$$${g}\left({z}\right)=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\:\sqrt{\mathrm{2}}{z}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:=\frac{{a}}{{z}}\:+\frac{{b}}{{z}−{z}_{\mathrm{1}} }\:+\frac{{c}}{{z}−{z}_{\mathrm{2}} } \\ $$$${a}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}{z}_{\mathrm{1}} {z}_{\mathrm{2}} }\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\:,{b}\:=\frac{{z}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}}{\:\sqrt{\mathrm{2}}{z}_{\mathrm{1}} \sqrt{\mathrm{2}}{i}\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }}\:=\frac{\mathrm{1}+{z}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{iz}_{\mathrm{1}} \sqrt{\mathrm{2}−{t}^{\mathrm{2}} }} \\ $$$${b}=\frac{{z}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}}{\:\sqrt{\mathrm{2}}{z}_{\mathrm{2}} \left(−\sqrt{\mathrm{2}}{i}\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{1}+{z}_{\mathrm{2}} ^{\mathrm{2}} }{−\mathrm{2}{iz}_{\mathrm{2}} \sqrt{\mathrm{2}−{t}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} {g}\left({z}\right){dz}\:=\int_{\mathrm{1}} ^{+\infty} \left(\frac{{a}}{{z}}+\frac{{b}}{{z}−{z}_{\mathrm{1}} }\:+\frac{{c}}{{z}−{z}_{\mathrm{2}} }\right){dz} \\ $$$$=\left[{aln}\mid{z}\mid\:+{bln}\mid{z}−{z}_{\mathrm{1}} \mid\:+{cln}\mid{z}−{z}_{\mathrm{2}} \mid\right]_{\mathrm{1}} ^{+\infty} \:{rest}\:{to}\:{simplify}\:{b}\:{and}\:{c} \\ $$$${be}\:{continued}… \\ $$
Answered by MJS last updated on 18/Apr/20
![∫(dx/((1+(√(x^2 +2)))^2 ))= [t=((x+(√(x^2 +2)))/( (√2))) → dx=((√(2(x^2 +2)))/(x+(√(x^2 +2))))dt] =(√2)∫((t^2 +1)/((t^2 +(√2)t+1)^2 ))dt= [Ostrogradski] =(((√2)t+2)/(t^2 +(√2)t+1))+2(√2)∫(dt/(t^2 +(√2)t+1))= =(((√2)t+2)/(t^2 +(√2)t+1))+4arctan ((√2)t+1) = =((x^2 +x+1−x(√(x^2 +2)))/(x^2 +1))+4arctan (x+1+(√(x^2 +2))) +C ⇒ ∫_0 ^∞ (dx/((1+(√(x^2 +2)))^2 ))=(π/2)−1](https://www.tinkutara.com/question/Q89721.png)
$$\int\frac{{dx}}{\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{dt}\right] \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=\frac{\sqrt{\mathrm{2}}{t}+\mathrm{2}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}+\mathrm{2}\sqrt{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}= \\ $$$$=\frac{\sqrt{\mathrm{2}}{t}+\mathrm{2}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}+\mathrm{4arctan}\:\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\:= \\ $$$$=\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}−{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{4arctan}\:\left({x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)\:+{C} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}}−\mathrm{1} \\ $$
Commented by turbo msup by abdo last updated on 18/Apr/20
$${thank}\:{you}\:{sir}. \\ $$