Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 125969 by nico last updated on 16/Dec/20

Commented by nico last updated on 16/Dec/20

help...

$${help}... \\ $$

Answered by MJS_new last updated on 16/Dec/20

let t=(x/( (√(1−x^2 )))) → dx=(1−x^2 )^(3/2) dt ⇒ ∫((t+1)/((t^2 +1)(t^2 −t+1)))dt=∫((t−1)/(t^2 +1))dt−∫((t−2)/(t^2 −t+1))dt= =(1/2)∫((2t)/(t^2 +1))dt−∫(dt/(t^2 +1))−(1/2)∫((2t−1)/(t^2 −t+1))dt+6∫(dt/((2t−1)^2 +3))= =(1/2)ln (t^2 +1) −arctan t −(1/2)ln (t^2 −t+1) +(√3)arctan (((√3)(2t−1))/3) = ... =(√3)arctan (((√3)(2x−(√(1−x^2 ))))/(3(√(1−x^2 )))) −arcsin x −(1/2)ln ∣1−x(√(1−x^2 ))∣ +C

$$\mathrm{let}\:{t}=\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\rightarrow\:{dx}=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} {dt} \\ $$$$\Rightarrow \\ $$$$\int\frac{{t}+\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{dt}=\int\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}−\int\frac{{t}−\mathrm{2}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}−\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}+\mathrm{6}\int\frac{{dt}}{\left(\mathrm{2}{t}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:−\mathrm{arctan}\:{t}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\:+\sqrt{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}−\mathrm{1}\right)}{\mathrm{3}}\:= \\ $$$$... \\ $$$$=\sqrt{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{\mathrm{3}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\mathrm{arcsin}\:{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{1}−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\mid\:+{C} \\ $$

Answered by mathmax by abdo last updated on 16/Dec/20

A =∫ ((x+(√(1−x^2 )))/(1−x(√(1−x^2 ))))dx we do ch. (√(1−x^2 ))=1−xt ⇒1−x^2 =(1−xt)^2 ⇒1−x^2 =1−2xt +x^2 t^2 ⇒−x^2 =−2xt+x^2 t^2 ⇒−x=−2t+xt^2 ⇒−2t+xt^2 +x =0 ⇒(1+t^2 )x=2t ⇒x=((2t)/(t^2 +1)) ⇒ (dx/dt) =((2(t^2 +1)−2t(2t))/((t^2 +1)^2 ))=((2t^2 +2−4t^2 )/((t^2 +1)^2 ))=((2−2t^2 )/((t^2 +1)^2 )) ⇒ A =∫ ((((2t)/(t^2 +1))+1−((2t^2 )/(t^2 +1)))/(1−((2t)/(t^2 +1))(1−((2t^2 )/(t^2 +1)))))×((2−2t^2 )/((t^2 +1)^2 ))dt =∫ ((2t+t^(2 ) +1−2t^2 )/((t^2 +1)^3 (1−((2t)/(t^2 +1))×((1−t^2 )/(t^2 +1)))))dt =∫ ((−t^2 +2t+1)/((t^2 +1)^3 (1−((2t−2t^3 )/((t^2 +1)^2 )))))dt =∫ ((−t^2 +2t+1)/((t^2 +1)( t^4 +2t^2 +1−2t+2t^3 )))dt =∫ ((−t^2 +2t+1)/((t^2 +1)(t^4 +2t^3 +2t^2 −2t+1)))dt rest decomposition of F(t)=((−t^2 +2t+1)/((t^2 +1)(t^4 +2t^3 +2t^2 −2t +1)))....be continued...

$$\mathrm{A}\:=\int\:\frac{\mathrm{x}+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}−\mathrm{x}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{ch}.\:\:\:\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }=\mathrm{1}−\mathrm{xt}\:\Rightarrow\mathrm{1}−\mathrm{x}^{\mathrm{2}} \:=\left(\mathrm{1}−\mathrm{xt}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{x}^{\mathrm{2}} \:=\mathrm{1}−\mathrm{2xt}\:+\mathrm{x}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \:\Rightarrow−\mathrm{x}^{\mathrm{2}} \:=−\mathrm{2xt}+\mathrm{x}^{\mathrm{2}} \:\mathrm{t}^{\mathrm{2}} \:\Rightarrow−\mathrm{x}=−\mathrm{2t}+\mathrm{xt}^{\mathrm{2}} \\ $$$$\Rightarrow−\mathrm{2t}+\mathrm{xt}^{\mathrm{2}} +\mathrm{x}\:=\mathrm{0}\:\Rightarrow\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{x}=\mathrm{2t}\:\Rightarrow\mathrm{x}=\frac{\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\frac{\mathrm{dx}}{\mathrm{dt}}\:=\frac{\mathrm{2}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{2t}\left(\mathrm{2t}\right)}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{2t}^{\mathrm{2}} \:+\mathrm{2}−\mathrm{4t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{2}−\mathrm{2t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{A}\:=\int\:\:\:\frac{\frac{\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}+\mathrm{1}−\frac{\mathrm{2t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{1}−\frac{\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\left(\mathrm{1}−\frac{\mathrm{2t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\right)}×\frac{\mathrm{2}−\mathrm{2t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\int\:\:\frac{\mathrm{2t}+\mathrm{t}^{\mathrm{2}\:} +\mathrm{1}−\mathrm{2t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}×\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\right)}\mathrm{dt} \\ $$$$=\int\:\:\frac{−\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}+\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{2t}−\mathrm{2t}^{\mathrm{3}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\right)}\mathrm{dt} \\ $$$$=\int\:\:\frac{−\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}+\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\:\:\mathrm{t}^{\mathrm{4}} +\mathrm{2t}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2t}+\mathrm{2t}^{\mathrm{3}} \right)}\mathrm{dt} \\ $$$$=\int\:\:\frac{−\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}+\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{4}} \:+\mathrm{2t}^{\mathrm{3}} \:+\mathrm{2t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}\right)}\mathrm{dt}\:\:\mathrm{rest}\:\mathrm{decomposition}\:\mathrm{of} \\ $$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{−\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}+\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{4}} \:+\mathrm{2t}^{\mathrm{3}} \:+\mathrm{2t}^{\mathrm{2}} −\mathrm{2t}\:+\mathrm{1}\right)}....\mathrm{be}\:\mathrm{continued}... \\ $$

Answered by liberty last updated on 17/Dec/20

let x^2 = (p^2 /(1+p^2 ))∧(1/x^2 )=((1+p^2 )/p^2 ) ⇒2x dx = ((2p(1+p^2 )−2p^3 )/((1+p^2 )^2 ))dp x dx = (p/((1+p^2 )^2 )) dp ⇒dx = (dp/((1+p^2 )^(3/2) )) I= ∫ (((p/( (√(1+p^2 ))))+1)/(1−(p/( (√(1+p^2 )))))) ((dp/((1+p^2 )^(3/2) ))) I=∫ ((p+(√(1+p^2 )))/( (√(1+p^2 ))−p))((dp/((1+p^2 )^(3/2) ))) let p = tan z I = ∫ (((tan z+sec z)/(sec z−tan z)))(((sec^2 z )/(sec^3 z)))dz I= ∫ (((sin z+1)/(1−sin z)))cos z dz I=∫(((1+2sin z+sin^2 z)/(cos z)))dz I=∫(((2+2sin z−cos^2 z)/(cos z)))dz = ∫(2sec z−cos z+((2sin z)/(cos z)))dz = 2ln ∣sec z+tan z∣−sin z−2ln ∣cos z ∣+c =2ln ∣((sec z+tan z)/(cos z)) ∣−sin z + c =2ln ∣((1+sin z)/(cos^2 z)) ∣−sin z + c =2ln ∣((1+(p/( (√(1+p^2 )))))/( (1/(1+p^2 )))) ∣−(p/( (√(1+p^2 )))) +c =2ln ∣1+p^2 +p(√(1+p^2 )) ∣−(p/( (√(1+p^2 )))) + c =−x+2ln ∣1+(x^2 /(1−x^2 ))+(x/(1−x^2 ))∣ +c =−x+2ln ∣(1/(1−x))∣ + c

$${let}\:{x}^{\mathrm{2}} =\:\frac{{p}^{\mathrm{2}} }{\mathrm{1}+{p}^{\mathrm{2}} }\wedge\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{\mathrm{1}+{p}^{\mathrm{2}} }{{p}^{\mathrm{2}} }\:\Rightarrow\mathrm{2}{x}\:{dx}\:=\:\frac{\mathrm{2}{p}\left(\mathrm{1}+{p}^{\mathrm{2}} \right)−\mathrm{2}{p}^{\mathrm{3}} }{\left(\mathrm{1}+{p}^{\mathrm{2}} \right)^{\mathrm{2}} }{dp} \\ $$$${x}\:{dx}\:=\:\frac{{p}}{\left(\mathrm{1}+{p}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dp}\:\Rightarrow{dx}\:=\:\frac{{dp}}{\left(\mathrm{1}+{p}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${I}=\:\int\:\frac{\frac{{p}}{\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}+\mathrm{1}}{\mathrm{1}−\frac{{p}}{\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}}\:\left(\frac{{dp}}{\left(\mathrm{1}+{p}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\right) \\ $$$${I}=\int\:\frac{{p}+\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }−{p}}\left(\frac{{dp}}{\left(\mathrm{1}+{p}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\right) \\ $$$${let}\:{p}\:=\:\mathrm{tan}\:{z}\: \\ $$$${I}\:=\:\int\:\left(\frac{\mathrm{tan}\:{z}+\mathrm{sec}\:{z}}{\mathrm{sec}\:{z}−\mathrm{tan}\:{z}}\right)\left(\frac{\mathrm{sec}\:^{\mathrm{2}} {z}\:}{\mathrm{sec}\:^{\mathrm{3}} {z}}\right){dz} \\ $$$${I}=\:\int\:\left(\frac{\mathrm{sin}\:{z}+\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:{z}}\right)\mathrm{cos}\:{z}\:{dz}\: \\ $$$${I}=\int\left(\frac{\mathrm{1}+\mathrm{2sin}\:{z}+\mathrm{sin}\:^{\mathrm{2}} {z}}{\mathrm{cos}\:{z}}\right){dz} \\ $$$${I}=\int\left(\frac{\mathrm{2}+\mathrm{2sin}\:{z}−\mathrm{cos}\:^{\mathrm{2}} {z}}{\mathrm{cos}\:{z}}\right){dz}\:=\:\int\left(\mathrm{2sec}\:{z}−\mathrm{cos}\:{z}+\frac{\mathrm{2sin}\:{z}}{\mathrm{cos}\:{z}}\right){dz} \\ $$$$=\:\mathrm{2ln}\:\mid\mathrm{sec}\:{z}+\mathrm{tan}\:{z}\mid−\mathrm{sin}\:{z}−\mathrm{2ln}\:\mid\mathrm{cos}\:{z}\:\mid+{c} \\ $$$$=\mathrm{2ln}\:\mid\frac{\mathrm{sec}\:{z}+\mathrm{tan}\:{z}}{\mathrm{cos}\:{z}}\:\mid−\mathrm{sin}\:{z}\:+\:{c}\: \\ $$$$=\mathrm{2ln}\:\mid\frac{\mathrm{1}+\mathrm{sin}\:{z}}{\mathrm{cos}\:^{\mathrm{2}} {z}}\:\mid−\mathrm{sin}\:{z}\:+\:{c}\: \\ $$$$=\mathrm{2ln}\:\mid\frac{\mathrm{1}+\frac{{p}}{\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}}{\:\frac{\mathrm{1}}{\mathrm{1}+{p}^{\mathrm{2}} }}\:\mid−\frac{{p}}{\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}\:+{c}\: \\ $$$$=\mathrm{2ln}\:\mid\mathrm{1}+{p}^{\mathrm{2}} +{p}\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }\:\mid−\frac{{p}}{\:\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }}\:+\:{c} \\ $$$$=−{x}+\mathrm{2ln}\:\mid\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\mid\:+{c} \\ $$$$=−{x}+\mathrm{2ln}\:\mid\frac{\mathrm{1}}{\mathrm{1}−{x}}\mid\:+\:{c}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com