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Question Number 60687 by maxmathsup by imad last updated on 24/May/19
calculate ∫ (((√(1+x^2 ))−2x)/( (√(1+x^2 )) +2x)) dx
$${calculate}\:\:\int\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{2}{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+\mathrm{2}{x}}\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 25/May/19
let A =∫ (((√(1+x^2 ))−2x)/( (√(1+x^2 )) +2x)) dx let use thechangement x =sh(t) ⇒ A =∫ ((ch(t)−2sh(t))/(ch(t)+2sh(t))) ch(t)dt = ∫ ((ch^2 t−sh(2t))/(ch(t)+2sh(t)))dt =∫ ((((1+ch(2t))/2)−sh(2t))/(ch(t)+2sh(t))) dt =(1/2) ∫ ((1+ch(2t)−2sh(2t))/(ch(t)+2sh(t))) dt 2A =∫ ((1+((e^(2t) +e^(−2t) )/2) −2((e^(2t) −e^(−2t) )/2))/(((e^t +e^(−t) )/2) +2 ((e^t −e^(−t) )/2))) dt = ∫ ((2 +e^(2t) +e^(−2t) −2 e^(2t) +2e^(−2t) )/(e^t +e^(−t) +2 e^t −2e^(−t) )) dt =_(e^t =u) ∫ ((2 −u^2 +3u^(−2) )/(3u−u^(−1) )) du =∫ ((2u^2 −u^4 +3)/(3u^3 −u)) du =− ∫ ((u^4 −2u^2 −3)/(3u^3 −u)) du let decompose F(u) =((u^4 −2u^2 −3)/(3u^3 −u)) ⇒ F(u) =(1/3) ((3u^4 −6u^2 −9)/(3u^3 −u)) ⇒ F(u) =(1/3) { ((3(u^3 −u )u −3u^2 −9)/(3u^3 −u))} =(1/3){ u −((3u^2 +9)/(3u^3 −u))} =(u/3) −((u^2 +3)/(3u^3 −u)) and g(u) =((u^2 +3)/(3u^3 −u)) =((u^2 +3)/(u(3u^2 −1)))=((u^2 +3)/(3u(u−(1/( (√3))))(u+(1/( (√3)))))) =(a/u) +(b/(u−(1/( (√3))))) +(c/(u+(1/( (√3))))) its eazh to find a ,b,c so ∫ F(u)du =aln∣u∣+bln∣u−(1/( (√3)))∣+c ln∣u+(1/( (√3)))∣ +c ⇒ A =−(a/2)ln∣u∣ −bln∣u−(1/( (√3)))∣−c ln∣u+(1/( (√3)))∣ +c u =e^t and t =argsh(x) =ln(x+(√(1+x^2 ))) ⇒ u =x+(√(1+x^2 )) ⇒ A =−(a/2)ln∣x+(√(1+x^2 ))∣ −bln∣x+(√(1+x^2 ))−(1/( (√3)))∣ −c ln∣x+(√(1+x^2 )) +(1/( (√3)))∣ +C .
$${let}\:{A}\:=\int\:\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{2}{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+\mathrm{2}{x}}\:{dx}\:\:{let}\:{use}\:{thechangement}\:{x}\:={sh}\left({t}\right)\:\Rightarrow \\ $$$${A}\:=\int\:\:\frac{{ch}\left({t}\right)−\mathrm{2}{sh}\left({t}\right)}{{ch}\left({t}\right)+\mathrm{2}{sh}\left({t}\right)}\:{ch}\left({t}\right){dt}\:=\:\int\:\:\:\frac{{ch}^{\mathrm{2}} {t}−{sh}\left(\mathrm{2}{t}\right)}{{ch}\left({t}\right)+\mathrm{2}{sh}\left({t}\right)}{dt} \\ $$$$=\int\:\:\frac{\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}−{sh}\left(\mathrm{2}{t}\right)}{{ch}\left({t}\right)+\mathrm{2}{sh}\left({t}\right)}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)−\mathrm{2}{sh}\left(\mathrm{2}{t}\right)}{{ch}\left({t}\right)+\mathrm{2}{sh}\left({t}\right)}\:{dt}\: \\ $$$$\mathrm{2}{A}\:=\int\:\:\:\frac{\mathrm{1}+\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\:−\mathrm{2}\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}}}{\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\:+\mathrm{2}\:\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}}\:{dt}\: \\ $$$$=\:\int\:\:\frac{\mathrm{2}\:+{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} \:−\mathrm{2}\:{e}^{\mathrm{2}{t}} \:+\mathrm{2}{e}^{−\mathrm{2}{t}} }{{e}^{{t}} +{e}^{−{t}} \:+\mathrm{2}\:{e}^{{t}} \:−\mathrm{2}{e}^{−{t}} }\:{dt}\:=_{{e}^{{t}} \:={u}} \:\:\:\:\int\:\:\:\frac{\mathrm{2}\:−{u}^{\mathrm{2}} \:+\mathrm{3}{u}^{−\mathrm{2}} }{\mathrm{3}{u}−{u}^{−\mathrm{1}} }\:{du} \\ $$$$=\int\:\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} −{u}^{\mathrm{4}} \:+\mathrm{3}}{\mathrm{3}{u}^{\mathrm{3}} −{u}}\:{du}\:\:\:=−\:\int\:\:\:\:\frac{{u}^{\mathrm{4}} \:−\mathrm{2}{u}^{\mathrm{2}} −\mathrm{3}}{\mathrm{3}{u}^{\mathrm{3}} −{u}}\:{du}\:\:{let}\:{decompose}\: \\ $$$${F}\left({u}\right)\:=\frac{{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} −\mathrm{3}}{\mathrm{3}{u}^{\mathrm{3}} −{u}}\:\Rightarrow\:{F}\left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\mathrm{3}{u}^{\mathrm{4}} −\mathrm{6}{u}^{\mathrm{2}} −\mathrm{9}}{\mathrm{3}{u}^{\mathrm{3}} −{u}}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}\:\left\{\:\:\:\frac{\mathrm{3}\left({u}^{\mathrm{3}} −{u}\:\right){u}\:−\mathrm{3}{u}^{\mathrm{2}} −\mathrm{9}}{\mathrm{3}{u}^{\mathrm{3}} −{u}}\right\}\:=\frac{\mathrm{1}}{\mathrm{3}}\left\{\:{u}\:−\frac{\mathrm{3}{u}^{\mathrm{2}} +\mathrm{9}}{\mathrm{3}{u}^{\mathrm{3}} −{u}}\right\} \\ $$$$=\frac{{u}}{\mathrm{3}}\:−\frac{{u}^{\mathrm{2}} \:+\mathrm{3}}{\mathrm{3}{u}^{\mathrm{3}} −{u}}\:\:{and}\:{g}\left({u}\right)\:=\frac{{u}^{\mathrm{2}} \:+\mathrm{3}}{\mathrm{3}{u}^{\mathrm{3}} −{u}}\:=\frac{{u}^{\mathrm{2}} \:+\mathrm{3}}{{u}\left(\mathrm{3}{u}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{{u}^{\mathrm{2}} \:+\mathrm{3}}{\mathrm{3}{u}\left({u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\left({u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)} \\ $$$$=\frac{{a}}{{u}}\:+\frac{{b}}{{u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\:+\frac{{c}}{{u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\:\:{its}\:{eazh}\:{to}\:{find}\:{a}\:,{b},{c}\:{so} \\ $$$$\int\:{F}\left({u}\right){du}\:={aln}\mid{u}\mid+{bln}\mid{u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mid+{c}\:{ln}\mid{u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mid\:+{c}\:\Rightarrow \\ $$$${A}\:=−\frac{{a}}{\mathrm{2}}{ln}\mid{u}\mid\:−{bln}\mid{u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mid−{c}\:{ln}\mid{u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mid\:+{c} \\ $$$${u}\:={e}^{{t}} \:{and}\:{t}\:={argsh}\left({x}\right)\:={ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\Rightarrow\right. \\ $$$${u}\:={x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\Rightarrow \\ $$$${A}\:=−\frac{{a}}{\mathrm{2}}{ln}\mid{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid\:−{bln}\mid{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mid\:−{c}\:{ln}\mid{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mid\:+{C}\:. \\ $$
Commented by aliesam last updated on 27/May/19
Answered by ajfour last updated on 24/May/19
x=cot θ I=−∫ ((cosec θ−2cot θ)/(cosec θ+2cot θ)) cosec^2 θdθ = −∫((1−2cos θ)/(sin^2 θ(1+2cos θ)))dθ = −∫((2sin^2 (θ/2) (dθ/2)×2)/(8sin^2 (θ/2)cos^4 (θ/2))) let (θ/2)= φ I=−(1/2)∫sec^2 φsec^2 φdφ =−(1/2){sec^2 φtan φ−∫2sec^2 φtan^2 φdφ} =−(1/2)sec^2 φtan φ−2I−tan φ+c I=−(1/6)sec^2 (θ/2)tan (θ/2)−(1/3)tan (θ/2)+c let tan (θ/2)=t and since tan θ=x ((2t)/(1−t^2 ))=x ⇒ xt^2 +2t−x=0 t=((−2±(√(4+4x^2 )))/(2x)) = −(1/x)±((√(1+x^2 ))/x) I=−(1/6)(1+t^2 )t−(t/3)+c .
$${x}=\mathrm{cot}\:\theta \\ $$$${I}=−\int\:\frac{\mathrm{cosec}\:\theta−\mathrm{2cot}\:\theta}{\mathrm{cosec}\:\theta+\mathrm{2cot}\:\theta}\:\mathrm{cosec}\:^{\mathrm{2}} \theta{d}\theta \\ $$$$\:\:=\:−\int\frac{\mathrm{1}−\mathrm{2cos}\:\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta\left(\mathrm{1}+\mathrm{2cos}\:\theta\right)}{d}\theta \\ $$$$\:\:\:=\:−\int\frac{\mathrm{2sin}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\:\frac{{d}\theta}{\mathrm{2}}×\mathrm{2}}{\mathrm{8sin}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\mathrm{cos}\:^{\mathrm{4}} \frac{\theta}{\mathrm{2}}} \\ $$$$\:{let}\:\frac{\theta}{\mathrm{2}}=\:\phi \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sec}\:^{\mathrm{2}} \phi\mathrm{sec}\:^{\mathrm{2}} \phi{d}\phi \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{sec}\:^{\mathrm{2}} \phi\mathrm{tan}\:\phi−\int\mathrm{2sec}\:^{\mathrm{2}} \phi\mathrm{tan}\:^{\mathrm{2}} \phi{d}\phi\right\} \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} \phi\mathrm{tan}\:\phi−\mathrm{2}{I}−\mathrm{tan}\:\phi+{c} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{sec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\mathrm{tan}\:\frac{\theta}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:\frac{\theta}{\mathrm{2}}+{c} \\ $$$${let}\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}={t}\:\:\:{and}\:{since}\:\mathrm{tan}\:\theta={x} \\ $$$$\:\:\:\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }={x}\:\:\Rightarrow\:\:{xt}^{\mathrm{2}} +\mathrm{2}{t}−{x}=\mathrm{0} \\ $$$$\:\:\:\:{t}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}{x}}\:=\:−\frac{\mathrm{1}}{{x}}\pm\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){t}−\frac{{t}}{\mathrm{3}}+{c}\:. \\ $$

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