Menu Close

Question-74720

Tinku Tara 0 Comments
Pin




Question Number 74720 by chess1 last updated on 29/Nov/19
Commented by mathmax by abdo last updated on 29/Nov/19
changement (√((x−1)/(x+1))) =t give x−1 =t^2 (x+1) ⇒(1−t^2 )x=1+t^2 ⇒ x=((1+t^2 )/(1−t^2 )) ⇒(dx/dt) =((2t(1−t^2 )−(1+t^2 )(−2t))/((1−t^2 )^2 )) =((2t−2t^3 +2t +2t^3 )/((t^2 −1)^2 )) =((4t)/((t^2 −1)^2 )) ⇒∫ (√((x−1)/(x+1)))dx =4∫(t^2 /((t^2 −1)^2 ))dt let decompose F(t)=(t^2 /((t^2 −1)^2 )) ⇒F(t)=(t^2 /((t+1)^2 (t−1)^2 )) =(a/(t+1)) +(b/((t+1)^2 )) +(c/(t−1)) +(d/((t−1)^2 )) F(−t)=F(t) ⇒((−a)/(t−1)) +(b/((t−1)^2 )) −(c/(t+1)) +(d/((t+1)^2 )) =(a/(t+1)) +(b/((t+1)^2 )) +(c/(t−1)) +(d/((t−1)^2 )) ⇒c=−a and b=d ⇒ F(t)=(a/(t+1)) +(b/((t+1)^2 ))−(a/(t−1)) +(b/((t−1)^2 )) b=(t+1)^2 F(t)∣_(t=−1) =(1/4) ⇒F(t)=(a/(t+1)) +(1/(4(t+1)^2 ))−(a/(t−1)) +(1/(4(t−1)^2 )) F(0)=0 =a+(1/4) +a+(1/4) =2a+(1/2) ⇒2a=−(1/2) ⇒a=−(1/4) ⇒ F(t)=(1/(4(t−1)))−(1/(4(t+1))) +(1/(4(t+1)^2 )) +(1/(4(t−1)^2 )) ⇒ ∫(√((x−1)/(x+1)))dx =∫ (dt/(t−1))−∫ (dt/(t+1)) +∫ (dt/((t+1)^2 )) +∫ (dt/((t−1)^2 )) =ln∣t−1∣−ln∣t+1∣−(1/(t+1))−(1/(t−1)) +C =ln∣((t−1)/(t+1))∣−((1/(t+1))+(1/(t−1))) +C =ln∣((t−1)/(t+1))∣−((2t)/(t^2 −1)) +C =ln∣(((√((x−1)/(x+1 )))−1)/( (√((x−1)/(x+1)))+1))∣−((2(√((x−1)/(x+1))))/(((x−1)/(x+1))−1)) +C =ln∣(((√((x−1)/(x+1)))−1)/( (√((x−1)/(x+1)))+1))∣ +(x+1)(√((x−1)/(x+1 ))) + C .
$${changement}\:\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:={t}\:{give}\:{x}−\mathrm{1}\:={t}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\:\Rightarrow\left(\mathrm{1}−{t}^{\mathrm{2}} \right){x}=\mathrm{1}+{t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow\frac{{dx}}{{dt}}\:=\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)−\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(−\mathrm{2}{t}\right)}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}\:+\mathrm{2}{t}^{\mathrm{3}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\int\:\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}{dx}\:=\mathrm{4}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\:{let}\:{decompose} \\ $$$${F}\left({t}\right)=\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{F}\left({t}\right)=\frac{{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)^{\mathrm{2}} \left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{b}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{{t}−\mathrm{1}}\:+\frac{{d}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(−{t}\right)={F}\left({t}\right)\:\Rightarrow\frac{−{a}}{{t}−\mathrm{1}}\:+\frac{{b}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{{c}}{{t}+\mathrm{1}}\:+\frac{{d}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{b}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$+\frac{{c}}{{t}−\mathrm{1}}\:+\frac{{d}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{c}=−{a}\:{and}\:{b}={d}\:\Rightarrow \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{b}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{{a}}{{t}−\mathrm{1}}\:+\frac{{b}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}=\left({t}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({t}\right)\mid_{{t}=−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow{F}\left({t}\right)=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({t}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{{a}}{{t}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0}\:={a}+\frac{\mathrm{1}}{\mathrm{4}}\:+{a}+\frac{\mathrm{1}}{\mathrm{4}}\:=\mathrm{2}{a}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{2}{a}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{4}\left({t}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left({t}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}{dx}\:=\int\:\:\frac{{dt}}{{t}−\mathrm{1}}−\int\:\frac{{dt}}{{t}+\mathrm{1}}\:+\int\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:+\int\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$={ln}\mid{t}−\mathrm{1}\mid−{ln}\mid{t}+\mathrm{1}\mid−\frac{\mathrm{1}}{{t}+\mathrm{1}}−\frac{\mathrm{1}}{{t}−\mathrm{1}}\:+{C} \\ $$$$={ln}\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid−\left(\frac{\mathrm{1}}{{t}+\mathrm{1}}+\frac{\mathrm{1}}{{t}−\mathrm{1}}\right)\:+{C}\:={ln}\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid−\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} −\mathrm{1}}\:+{C} \\ $$$$={ln}\mid\frac{\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}\:}}−\mathrm{1}}{\:\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}+\mathrm{1}}\mid−\frac{\mathrm{2}\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}}{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}−\mathrm{1}}\:+{C} \\ $$$$={ln}\mid\frac{\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}−\mathrm{1}}{\:\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}+\mathrm{1}}\mid\:+\left({x}+\mathrm{1}\right)\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}\:}}\:+\:{C}\:. \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Nov/19
we can simplify we get ∫ (√((x−1)/(x+1)))dx =ln∣(((√(x−1))−(√(x+1)))/( (√(x−1)) +(√(x+1))))∣+(√(x^2 −1)) +C .
$${we}\:{can}\:{simplify}\:{we}\:{get}\: \\ $$$$\int\:\:\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}{dx}\:={ln}\mid\frac{\sqrt{{x}−\mathrm{1}}−\sqrt{{x}+\mathrm{1}}}{\:\sqrt{{x}−\mathrm{1}}\:+\sqrt{{x}+\mathrm{1}}}\mid+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:+{C}\:. \\ $$
Commented by chess1 last updated on 30/Nov/19
thanks
$$\mathrm{thanks} \\ $$
Commented by mathmax by abdo last updated on 30/Nov/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$
Commented by mathmax by abdo last updated on 06/Dec/19
another way we use the changement x =ch(2t) ⇒ I =∫(√((ch(2t)−1)/(ch(2t)+1)))2sh(2t)dt =2∫ (√((2sh^2 (t))/(2ch^2 (t))))ch(2t)dt =2 ∫ th(t)2sh(t)ch(t)dt =4 ∫ ((sh(t))/(ch(t)))sh(t)ch(t)dt =4 ∫ sh^2 t dt =4 ∫((ch(2t)−1)/2)dt =2 ∫ ch(2t)dt−2t =sh(2t)−2t +c =(√(ch^2 (2t)−1)) −argch(x) +c =(√(x^2 −1)) −ln(x+(√(x^2 −1))) +c
$${another}\:{way}\:\:{we}\:{use}\:{the}\:{changement}\:{x}\:={ch}\left(\mathrm{2}{t}\right)\:\Rightarrow \\ $$$${I}\:=\int\sqrt{\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{{ch}\left(\mathrm{2}{t}\right)+\mathrm{1}}}\mathrm{2}{sh}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\int\:\:\sqrt{\frac{\mathrm{2}{sh}^{\mathrm{2}} \left({t}\right)}{\mathrm{2}{ch}^{\mathrm{2}} \left({t}\right)}}{ch}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int\:\:{th}\left({t}\right)\mathrm{2}{sh}\left({t}\right){ch}\left({t}\right){dt}\:=\mathrm{4}\:\int\:\:\frac{{sh}\left({t}\right)}{{ch}\left({t}\right)}{sh}\left({t}\right){ch}\left({t}\right){dt} \\ $$$$=\mathrm{4}\:\int\:{sh}^{\mathrm{2}} {t}\:{dt}\:=\mathrm{4}\:\int\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}{dt}\:=\mathrm{2}\:\int\:{ch}\left(\mathrm{2}{t}\right){dt}−\mathrm{2}{t} \\ $$$$={sh}\left(\mathrm{2}{t}\right)−\mathrm{2}{t}\:+{c}\:\:=\sqrt{{ch}^{\mathrm{2}} \left(\mathrm{2}{t}\right)−\mathrm{1}}\:−{argch}\left({x}\right)\:+{c} \\ $$$$=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:−{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:+{c} \\ $$
Answered by Tanmay chaudhury last updated on 29/Nov/19
∫((x−1)/( (√(x^2 −1))))dx (1/2)∫((d(x^2 −1))/((x^2 −1)^(1/2) ))−∫(dx/( (√(x^2 −1)))) I_1 =(1/2)×(((x^2 −1)^(((−1)/2)+1) )/(((−1)/2)+1))+c_1 ⇛(x^2 −1)^(1/2) +c_1 I_2 =∫(dx/( (√(x^2 −1)))) let x=secα dx=secαtanαdα =∫((secαtanαdα)/(tanα))=∫secαdα=ln(secα+tanα)+c_2 =ln(x+(√(x^2 −1)) )+c_2
$$\int\frac{{x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }−\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}+\mathrm{1}} }{\frac{−\mathrm{1}}{\mathrm{2}}+\mathrm{1}}+{c}_{\mathrm{1}} \Rrightarrow\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\:\:{let}\:{x}={sec}\alpha\:\:{dx}={sec}\alpha{tan}\alpha{d}\alpha \\ $$$$=\int\frac{{sec}\alpha{tan}\alpha{d}\alpha}{{tan}\alpha}=\int{sec}\alpha{d}\alpha={ln}\left({sec}\alpha+{tan}\alpha\right)+{c}_{\mathrm{2}} \\ $$$$={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\right)+{c}_{\mathrm{2}} \\ $$
Commented by chess1 last updated on 30/Nov/19
thanks
$$\mathrm{thanks} \\ $$
Answered by Kunal12588 last updated on 30/Nov/19
I=∫(√((x−1)/(x+1)))dx=∫((x−1)/( (√(x^2 −1))))dx =∫(x/( (√(x^2 −1))))dx−∫(dx/( (√(x^2 −1)))) =(1/2)∫((2x)/( (√(x^2 −1))))dx−log∣x+(√(x^2 −1))∣+C let x^2 −1=t⇒2xdx=dt I=(1/2)∫(dt/( (√t)))−log∣x+(√(x^2 −1))∣+C =(1/2)(((√t)/(1/2)))−log∣x+(√(x^2 −1))∣+C =(√(x^2 −1))−log∣x+(√(x^2 −1))∣+C
$${I}=\int\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}{dx}=\int\frac{{x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx} \\ $$$$=\int\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx}−\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx}−{log}\mid{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\mid+{C} \\ $$$${let}\:{x}^{\mathrm{2}} −\mathrm{1}={t}\Rightarrow\mathrm{2}{xdx}={dt} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\:\sqrt{{t}}}−{log}\mid{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\sqrt{{t}}}{\mathrm{1}/\mathrm{2}}\right)−{log}\mid{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\mid+{C} \\ $$$$=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−{log}\mid{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\mid+{C} \\ $$

Pin

Leave a Reply

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