Menu Close

dx-x-x-2-2x-1-tan-1-x-1-x-2-2x-1-C-I-m-confused-in-choosing-the-right-substitution-so-that-we-can-get-the-result-above-Please-help-

Tinku Tara 0 Comments
Pin




Question Number 46369 by Joel578 last updated on 24/Oct/18
∫ (dx/(x(√(x^2 + 2x − 1)))) = tan^(−1) (((x − 1)/( (√(x^2 + 2x − 1))))) + C I′m confused in choosing the right substitution so that we can get the result above. Please help...
$$\int\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{1}}}\:\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}\:−\:\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{1}}}\right)\:+\:{C} \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{confused}\:\mathrm{in}\:\mathrm{choosing}\:\mathrm{the}\:\mathrm{right}\:\mathrm{substitution} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mathrm{the}\:\mathrm{result}\:\mathrm{above}. \\ $$$$\mathrm{Please}\:\mathrm{help}… \\ $$
Commented by maxmathsup by imad last updated on 24/Oct/18
let A =∫ (dx/(x(√(x^2 +2x−1)))) ⇒A=∫ (dx/(x(√(x^2 +2x+1−2)))) =∫ (dx/(x(√((x+1)^2 −2)))) changement x+1=(√2)ch(t)give A = ∫ (((√2)sh(t))/(((√2)ch(t)−1)(√(2s))h(t)))dt = ∫ (dt/( (√2)(((e^t +e^(−t) )/2))−1)) = ∫ ((2dt)/( (√2)(e^t +e^(−t) )−2)) =_(e^t =u) ∫ (2/( (√2)(u+u^(−1) )−2)) (du/u) = ∫ ((2du)/( (√2)u^2 −2u +(√2))) =(√2)∫ (du/(u^2 −(√2)u +1)) roots of u^2 −(√2)u +1 →Δ^′ =2−1=1 ⇒u_1 =1+(√2) and u_2 =(√2)−1 ⇒ A=(√2)∫ (du/((u−(1+(√2))(u−((√2)−1)))) =((√2)/2)∫ {(1/(u−((√2)+1))) −(1/(u−((√2)−1)))}du =((√2)/2)ln∣((u−((√2)+1))/(u−((√2)−1)))∣ +c ⇒ A =((√2)/2)ln∣((e^t −(√2)−1)/(u−(√2)+1))∣ +c but t=argch(((x+1)/( (√2))))=ln(((x+1)/( (√2))) +(√((((x+1)/( (√2))))^2 −1))) ⇒ A =((√2)/2)ln∣((((x+1)/( (√2)))+(√((((x+1)/( (√2) )))^2 −1))−(√2)−1)/(((x+1)/( (√2)))+(√((((x+1)/( (√2))))^2 −1))−(√2)+1))∣ +c .
$${let}\:{A}\:=\int\:\:\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}\:\Rightarrow{A}=\int\:\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}−\mathrm{2}}}\:=\int\:\:\frac{{dx}}{{x}\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}}} \\ $$$${changement}\:{x}+\mathrm{1}=\sqrt{\mathrm{2}}{ch}\left({t}\right){give}\: \\ $$$${A}\:=\:\int\:\:\:\frac{\sqrt{\mathrm{2}}{sh}\left({t}\right)}{\left(\sqrt{\mathrm{2}}{ch}\left({t}\right)−\mathrm{1}\right)\sqrt{\mathrm{2}{s}}{h}\left({t}\right)}{dt}\:=\:\int\:\:\:\:\frac{{dt}}{\:\sqrt{\mathrm{2}}\left(\frac{{e}^{{t}} +{e}^{−{t}} }{\mathrm{2}}\right)−\mathrm{1}} \\ $$$$=\:\int\:\:\:\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{2}}\left({e}^{{t}} \:+{e}^{−{t}} \right)−\mathrm{2}}\:\:=_{{e}^{{t}} ={u}} \:\:\:\:\:\:\int\:\:\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}\left({u}+{u}^{−\mathrm{1}} \right)−\mathrm{2}}\:\frac{{du}}{{u}} \\ $$$$=\:\int\:\:\:\:\:\frac{\mathrm{2}{du}}{\:\sqrt{\mathrm{2}}{u}^{\mathrm{2}} −\mathrm{2}{u}\:+\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}}\int\:\:\:\frac{{du}}{{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}{u}\:+\mathrm{1}} \\ $$$${roots}\:{of}\:{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}{u}\:+\mathrm{1}\:\rightarrow\Delta^{'} =\mathrm{2}−\mathrm{1}=\mathrm{1}\:\Rightarrow{u}_{\mathrm{1}} =\mathrm{1}+\sqrt{\mathrm{2}}\:\:{and}\:{u}_{\mathrm{2}} =\sqrt{\mathrm{2}}−\mathrm{1}\:\Rightarrow \\ $$$${A}=\sqrt{\mathrm{2}}\int\:\:\:\frac{{du}}{\left({u}−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left({u}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right)\right.} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\:\:\:\left\{\frac{\mathrm{1}}{{u}−\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{{u}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}\right\}{du} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{ln}\mid\frac{{u}−\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{{u}−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}\mid\:+{c}\:\Rightarrow \\ $$$${A}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{ln}\mid\frac{{e}^{{t}} \:−\sqrt{\mathrm{2}}−\mathrm{1}}{{u}−\sqrt{\mathrm{2}}+\mathrm{1}}\mid\:+{c}\:\:{but}\:{t}={argch}\left(\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)={ln}\left(\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\sqrt{\left(\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow \\ $$$${A}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{ln}\mid\frac{\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\sqrt{\left(\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\right)^{\mathrm{2}} −\mathrm{1}}−\sqrt{\mathrm{2}}−\mathrm{1}}{\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\sqrt{\left(\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\mathrm{1}}−\sqrt{\mathrm{2}}+\mathrm{1}}\mid\:+{c}\:. \\ $$$$ \\ $$
Answered by Smail last updated on 24/Oct/18
I=∫(dx/(x(√(x^2 +2x−1))))=∫(dx/(x(√((x+1)^2 −2)))) =∫(dx/(x(√(2((((x+1)/( (√2))))^2 −1)))))=(1/( (√2)))∫(dx/(x(√((((x+1)/( (√2))))^2 −1)))) t=((x+1)/( (√2)))⇒dx=(√2)dt I=∫(dt/(((√2)t−1)(√(t^2 −1)))) cosh(u)=t⇒dt=sinh(u)du I=∫(du/(((√2)cosh(u)−1)))=∫(du/( (√2)(((e^u +e^(−u) )/2))−1)) =2∫(du/( (√2)(e^u +e^(−u) )−2))=2∫((e^u du)/( (√2)e^(2u) +(√2)−2e^u )) z=e^u ⇒dz=e^u du I=2∫(dz/( (√2)z^2 −2z+(√2)))=(√2)∫(dz/(z^2 −(√2)z+1)) =(√2)∫(dz/(z^2 −2×((√2)/2)z+(1/2)−(1/2)+1)) =(√2)∫(dz/((z−((√2)/2))^2 +(1/2)))=(√2)∫(dz/(((((√2)z−1)/( (√2))))^2 +(1/2))) =(√2)∫(dz/((1/2)([(√2)((((√2)z−1)/( (√2))))]^2 +1))) =2(√2)∫(dz/(((√2)z−1)^2 +1)) v=(√2)z−1⇒dv=(√2)dz I=2∫(dv/(v^2 +1))=2tan^(−1) (v)+C =2tan^(−1) ((√2)z−1)+C=2tan^(−1) ((√2)e^u −1)+C =2tan^(−1) ((√2)e^(cosh^(−1) (t)) −1)+C =2tan^(−1) ((√2)(t+(√(t^2 −1)))−1)+C =2tan^(−1) ((√2)(((x+1)/( (√2)))+(√((((x+1)^2 )/2)−1)))−1)+C =2tan^(−1) (x+1+(√(x^2 +2x−1))−1)+C ∫(dx/(x(√(x^2 +2x−1))))=2tan^(−1) (x+(√(x^2 +2x−1)))+C
$${I}=\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}=\int\frac{{dx}}{{x}\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}}} \\ $$$$=\int\frac{{dx}}{{x}\sqrt{\mathrm{2}\left(\left(\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\mathrm{1}\right)}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dx}}{{x}\sqrt{\left(\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\mathrm{1}}} \\ $$$${t}=\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\Rightarrow{dx}=\sqrt{\mathrm{2}}{dt} \\ $$$${I}=\int\frac{{dt}}{\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${cosh}\left({u}\right)={t}\Rightarrow{dt}={sinh}\left({u}\right){du} \\ $$$${I}=\int\frac{{du}}{\left(\sqrt{\mathrm{2}}{cosh}\left({u}\right)−\mathrm{1}\right)}=\int\frac{{du}}{\:\sqrt{\mathrm{2}}\left(\frac{{e}^{{u}} +{e}^{−{u}} }{\mathrm{2}}\right)−\mathrm{1}} \\ $$$$=\mathrm{2}\int\frac{{du}}{\:\sqrt{\mathrm{2}}\left({e}^{{u}} +{e}^{−{u}} \right)−\mathrm{2}}=\mathrm{2}\int\frac{{e}^{{u}} {du}}{\:\sqrt{\mathrm{2}}{e}^{\mathrm{2}{u}} +\sqrt{\mathrm{2}}−\mathrm{2}{e}^{{u}} } \\ $$$${z}={e}^{{u}} \Rightarrow{dz}={e}^{{u}} {du} \\ $$$${I}=\mathrm{2}\int\frac{{dz}}{\:\sqrt{\mathrm{2}}{z}^{\mathrm{2}} −\mathrm{2}{z}+\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}\int\frac{{dz}}{{z}^{\mathrm{2}} −\sqrt{\mathrm{2}}{z}+\mathrm{1}} \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{dz}}{{z}^{\mathrm{2}} −\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{z}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}} \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{dz}}{\left({z}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}=\sqrt{\mathrm{2}}\int\frac{{dz}}{\left(\frac{\sqrt{\mathrm{2}}{z}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{dz}}{\frac{\mathrm{1}}{\mathrm{2}}\left(\left[\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}{z}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right]^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\int\frac{{dz}}{\left(\sqrt{\mathrm{2}}{z}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${v}=\sqrt{\mathrm{2}}{z}−\mathrm{1}\Rightarrow{dv}=\sqrt{\mathrm{2}}{dz} \\ $$$${I}=\mathrm{2}\int\frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2}{tan}^{−\mathrm{1}} \left({v}\right)+{C} \\ $$$$=\mathrm{2}{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}{z}−\mathrm{1}\right)+{C}=\mathrm{2}{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}{e}^{{u}} −\mathrm{1}\right)+{C} \\ $$$$=\mathrm{2}{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}{e}^{{cosh}^{−\mathrm{1}} \left({t}\right)} −\mathrm{1}\right)+{C} \\ $$$$=\mathrm{2}{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)−\mathrm{1}\right)+{C} \\ $$$$=\mathrm{2}{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\left(\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\sqrt{\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}}\right)−\mathrm{1}\right)+{C} \\ $$$$=\mathrm{2}{tan}^{−\mathrm{1}} \left({x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}−\mathrm{1}\right)+{C} \\ $$$$\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}=\mathrm{2}{tan}^{−\mathrm{1}} \left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}\right)+{C} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Oct/18
t=(1/x) x=(1/t) dx=((−dt)/t^2 ) ∫((−dt)/(t^2 ×(1/t)×(√((1/t^2 )+(2/t)−1)) )) =−1×∫(dt/(t(√((1+2t−t^2 )/t^2 )))) =−1×∫(dt/( (√(1−(t^2 −2t+1−1))))) =−1×∫(dt/( (√(2−(t−1)^2 )) )) =−1×sin^(−1) (((t−1)/( (√2))))+c =−sim^(−1) ((((1/x)−1)/( (√2))))+c =−tan^(−1) (((t−1)/( (√(2−(t−1)^2 ))))) =−tan^(−1) ((((1/x)−1)/( (√(2−(1/x^2 )+(2/x)−1))))) =−tan^(−1) (((1−x)/( (√(x^2 −1+2x))))) =tan^(−1) (((x−1)/( (√(x^2 +2x−1)))))+c
$${t}=\frac{\mathrm{1}}{{x}}\:\:\:{x}=\frac{\mathrm{1}}{{t}}\:\:{dx}=\frac{−{dt}}{{t}^{\mathrm{2}} } \\ $$$$\int\frac{−{dt}}{{t}^{\mathrm{2}} ×\frac{\mathrm{1}}{{t}}×\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\frac{\mathrm{2}}{{t}}−\mathrm{1}}\:} \\ $$$$=−\mathrm{1}×\int\frac{{dt}}{{t}\sqrt{\frac{\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} }}} \\ $$$$=−\mathrm{1}×\int\frac{{dt}}{\:\sqrt{\mathrm{1}−\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}−\mathrm{1}\right)}} \\ $$$$=−\mathrm{1}×\int\frac{{dt}}{\:\sqrt{\mathrm{2}−\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:} \\ $$$$=−\mathrm{1}×{sin}^{−\mathrm{1}} \left(\frac{{t}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$$$=−{sim}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{{x}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+{c}\: \\ $$$$=−{tan}^{−\mathrm{1}} \left(\frac{{t}−\mathrm{1}}{\:\sqrt{\mathrm{2}−\left({t}−\mathrm{1}\right)^{\mathrm{2}} }}\right) \\ $$$$=−{tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{{x}}−\mathrm{1}}{\:\sqrt{\mathrm{2}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{2}}{{x}}−\mathrm{1}}}\right) \\ $$$$=−{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}{x}}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}}}\right)+{c} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *