Menu Close

2-9x-2-x-dx-SOS-SOS-HELP-

Tinku Tara 0 Comments
Pin




Question Number 141496 by cesarL last updated on 19/May/21
∫((√(2+9x^2 ))/x)dx SOS SOS HELP
$$\int\frac{\sqrt{\mathrm{2}+\mathrm{9}{x}^{\mathrm{2}} }}{{x}}{dx} \\ $$$${SOS}\:{SOS}\:{HELP} \\ $$
Answered by qaz last updated on 19/May/21
∫(√(2+9x^2 ))(dx/x) =(1/2)∫(√(2+9x^2 ))((d(x^2 ))/x^2 ) =(1/2)∫(√(2+9y))(dy/y)................x^2 =y =(1/2)∫(√(2(1+(9/2)y)))(dy/y) =(√2)∫sec^3 t∙(dt/(tan t))..............(9/2)y=tan^2 t.....dy=(4/9)tan tsec^2 tdt =(√2)∫(dt/(cos^2 tsin t))..............cos t=((√2)/( (√(9y+2)))) =(√2)∫((d(cos t))/((cos^2 t−1)cos^2 t)) =(√2)∫((1/(cos^2 t−1))−(1/(cos^2 t)))d(cos t) =(1/( (√2)))ln∣((cos t−1)/(cos t+1))∣+((√2)/(cos t))+C =(1/( (√2)))ln∣((((√2)/( (√(9x^2 +2))))−1)/(((√2)/( (√(9x^2 +2))))+1))∣+(√(9x^2 +2))+C
$$\int\sqrt{\mathrm{2}+\mathrm{9}{x}^{\mathrm{2}} }\frac{{dx}}{{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\mathrm{2}+\mathrm{9}{x}^{\mathrm{2}} }\frac{{d}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\mathrm{2}+\mathrm{9}{y}}\frac{{dy}}{{y}}…………….{x}^{\mathrm{2}} ={y} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{9}}{\mathrm{2}}{y}\right)}\frac{{dy}}{{y}} \\ $$$$=\sqrt{\mathrm{2}}\int\mathrm{sec}\:^{\mathrm{3}} {t}\centerdot\frac{{dt}}{\mathrm{tan}\:{t}}…………..\frac{\mathrm{9}}{\mathrm{2}}{y}=\mathrm{tan}^{\mathrm{2}} \:{t}…..{dy}=\frac{\mathrm{4}}{\mathrm{9}}\mathrm{tan}\:{t}\mathrm{sec}\:^{\mathrm{2}} {tdt} \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{dt}}{\mathrm{cos}\:^{\mathrm{2}} {t}\mathrm{sin}\:{t}}…………..\mathrm{cos}\:{t}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{9}{y}+\mathrm{2}}} \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{d}\left(\mathrm{cos}\:{t}\right)}{\left(\mathrm{cos}\:^{\mathrm{2}} {t}−\mathrm{1}\right)\mathrm{cos}\:^{\mathrm{2}} {t}} \\ $$$$=\sqrt{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {t}}\right){d}\left(\mathrm{cos}\:{t}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\mid\frac{\mathrm{cos}\:{t}−\mathrm{1}}{\mathrm{cos}\:{t}+\mathrm{1}}\mid+\frac{\sqrt{\mathrm{2}}}{\mathrm{cos}\:{t}}+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\mid\frac{\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{9}{x}^{\mathrm{2}} +\mathrm{2}}}−\mathrm{1}}{\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{9}{x}^{\mathrm{2}} +\mathrm{2}}}+\mathrm{1}}\mid+\sqrt{\mathrm{9}{x}^{\mathrm{2}} +\mathrm{2}}+{C} \\ $$
Answered by MJS_new last updated on 19/May/21
∫((√(a^2 x^2 +b^2 ))/x)dx= [x=((b(t^2 −1))/(2at)) ⇔ t=((ax+(√(a^2 x^2 +b^2 )))/b) → dx=((b(√(a^2 x^2 +b^2 )))/(a(ax+(√(a^2 x^2 +b^2 )))))dt] =(b/2)∫(((t^2 +1)^2 )/(t^2 (t^2 −1)))dt= =(b/2)∫((t^2 −1)/t^2 )dt+b∫(dt/(t−1))−b∫(dt/(t+1))= =((b(t^2 +1))/(2t))+bln ((t−1)/(t+1))= =(√(a^2 x^2 +b^2 ))+bln ∣((b−(√(a^2 x^2 +b^2 )))/x)∣ +C
$$\int\frac{\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{x}=\frac{{b}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{at}}\:\Leftrightarrow\:{t}=\frac{{ax}+\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{b}}\:\rightarrow\:{dx}=\frac{{b}\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{a}\left({ax}+\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}{dt}\right] \\ $$$$=\frac{{b}}{\mathrm{2}}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)}{dt}= \\ $$$$=\frac{{b}}{\mathrm{2}}\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} }{dt}+{b}\int\frac{{dt}}{{t}−\mathrm{1}}−{b}\int\frac{{dt}}{{t}+\mathrm{1}}= \\ $$$$=\frac{{b}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}{t}}+{b}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}= \\ $$$$=\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} }+{b}\mathrm{ln}\:\mid\frac{{b}−\sqrt{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{x}}\mid\:+{C} \\ $$
Answered by mathmax by abdo last updated on 20/May/21
Ψ=∫ ((√(2+9x^2 ))/x)dx ⇒Ψ=_(3x=(√2)sht) 3 ∫ (((√2)ch(t))/( (√2)sht))×((√2)/3)cht dt =(√2)∫ ((ch^2 t)/(sht))dt =((√2)/2)∫ ((1+ch(2t))/(sht))dt =((√2)/2)∫ ((1+((e^(2t) +e^(−2t) )/2))/((e^t −e^(−t) )/2))dt =((√2)/2)∫ ((2+e^(2t) +e^(−2t) )/(e^t −e^(−t) ))dt =_(e^t =y) (1/( (√2)))∫ ((2+y^2 +y^(−2) )/(y−y^(−1) ))(dy/y) ⇒(√2)Ψ=∫ ((y^2 +y^(−2) +2)/(y^2 −1))dy =∫ ((y^4 +1+2y^2 )/(y^2 (y^2 −1)))dy =∫ ((y^4 +2y^2 +1)/(y^4 −y^2 ))dy =∫ ((y^4 −y^2 +3y^2 +1)/(y^4 −y^2 ))dy =y +∫ ((3y^2 +1)/(y^2 (y^2 −1)))dy let decompose F(y)=((3y^2 +1)/(y^2 (y^2 −1)))=((3y^2 +1)/(y^2 (y−1)(y+1))) F(y)=(a/y)+(b/y^2 ) +(c/(y−1)) +(d/(y+1)) b=−1 , c=(4/2)=2 , d=(4/(−2))=−2 ⇒F(y)=(a/y)−(1/y^2 )+(2/(y−1))−(2/(y+1)) F(2)=((13)/(12)) =(a/2)−(1/4) +2−(2/3) =(a/2)−(1/4)+(4/3) =(a/2)+((13)/(12)) ⇒a=0 ⇒ ∫ F(y)dy =(1/y) +2log∣((y−1)/(y+1))∣ +C =e^(−t) +2log∣((e^t −1)/(e^t +1))∣ +C but t=argsh(((3x)/( (√2)))) =log(((3x)/( (√2)))+(√(1+((9x^2 )/2)))) ⇒ (√2)Ψ=((3x)/( (√2)))+(√(1+((9x^2 )/2)))+2log∣((((3x)/( (√2)))+(√(1+((9x^2 )/2)))−1)/(((3x)/( (√2)))+(√(1+((9x^2 )/2)))+1))∣ +C ⇒ Ψ =(1/2)(3x+(√(2+9x^2 ))) +(√2)log∣((3x+(√(2+9x^2 ))−(√2))/(3x+(√(2+9x^2 ))+(√2)))∣ +C
$$\Psi=\int\:\frac{\sqrt{\mathrm{2}+\mathrm{9x}^{\mathrm{2}} }}{\mathrm{x}}\mathrm{dx}\:\:\Rightarrow\Psi=_{\mathrm{3x}=\sqrt{\mathrm{2}}\mathrm{sht}} \:\:\mathrm{3}\:\int\:\:\frac{\sqrt{\mathrm{2}}\mathrm{ch}\left(\mathrm{t}\right)}{\:\sqrt{\mathrm{2}}\mathrm{sht}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\mathrm{cht}\:\mathrm{dt} \\ $$$$=\sqrt{\mathrm{2}}\int\:\frac{\mathrm{ch}^{\mathrm{2}} \mathrm{t}}{\mathrm{sht}}\mathrm{dt}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\:\:\frac{\mathrm{1}+\mathrm{ch}\left(\mathrm{2t}\right)}{\mathrm{sht}}\mathrm{dt} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\:\:\frac{\mathrm{1}+\frac{\mathrm{e}^{\mathrm{2t}} +\mathrm{e}^{−\mathrm{2t}} }{\mathrm{2}}}{\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }{\mathrm{2}}}\mathrm{dt}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\:\:\:\frac{\mathrm{2}+\mathrm{e}^{\mathrm{2t}} \:+\mathrm{e}^{−\mathrm{2t}} }{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }\mathrm{dt} \\ $$$$=_{\mathrm{e}^{\mathrm{t}} \:=\mathrm{y}} \:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:\:\frac{\mathrm{2}+\mathrm{y}^{\mathrm{2}} \:+\mathrm{y}^{−\mathrm{2}} }{\mathrm{y}−\mathrm{y}^{−\mathrm{1}} }\frac{\mathrm{dy}}{\mathrm{y}}\:\Rightarrow\sqrt{\mathrm{2}}\Psi=\int\:\:\frac{\mathrm{y}^{\mathrm{2}} \:+\mathrm{y}^{−\mathrm{2}} +\mathrm{2}}{\mathrm{y}^{\mathrm{2}} −\mathrm{1}}\mathrm{dy} \\ $$$$=\int\:\:\frac{\mathrm{y}^{\mathrm{4}} \:+\mathrm{1}+\mathrm{2y}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right)}\mathrm{dy}\:\:=\int\:\frac{\mathrm{y}^{\mathrm{4}} \:+\mathrm{2y}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{y}^{\mathrm{4}} −\mathrm{y}^{\mathrm{2}} }\mathrm{dy} \\ $$$$=\int\:\frac{\mathrm{y}^{\mathrm{4}} −\mathrm{y}^{\mathrm{2}} \:+\mathrm{3y}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{y}^{\mathrm{4}} −\mathrm{y}^{\mathrm{2}} }\mathrm{dy}\:=\mathrm{y}\:+\int\:\frac{\mathrm{3y}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right)}\mathrm{dy} \\ $$$$\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{y}\right)=\frac{\mathrm{3y}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{\mathrm{3y}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{y}^{\mathrm{2}} \left(\mathrm{y}−\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)} \\ $$$$\mathrm{F}\left(\mathrm{y}\right)=\frac{\mathrm{a}}{\mathrm{y}}+\frac{\mathrm{b}}{\mathrm{y}^{\mathrm{2}} }\:+\frac{\mathrm{c}}{\mathrm{y}−\mathrm{1}}\:+\frac{\mathrm{d}}{\mathrm{y}+\mathrm{1}} \\ $$$$\mathrm{b}=−\mathrm{1}\:,\:\mathrm{c}=\frac{\mathrm{4}}{\mathrm{2}}=\mathrm{2}\:\:,\:\mathrm{d}=\frac{\mathrm{4}}{−\mathrm{2}}=−\mathrm{2}\:\Rightarrow\mathrm{F}\left(\mathrm{y}\right)=\frac{\mathrm{a}}{\mathrm{y}}−\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{y}−\mathrm{1}}−\frac{\mathrm{2}}{\mathrm{y}+\mathrm{1}} \\ $$$$\mathrm{F}\left(\mathrm{2}\right)=\frac{\mathrm{13}}{\mathrm{12}}\:=\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\:+\mathrm{2}−\frac{\mathrm{2}}{\mathrm{3}}\:=\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{4}}{\mathrm{3}}\:=\frac{\mathrm{a}}{\mathrm{2}}+\frac{\mathrm{13}}{\mathrm{12}}\:\Rightarrow\mathrm{a}=\mathrm{0}\:\Rightarrow \\ $$$$\int\:\mathrm{F}\left(\mathrm{y}\right)\mathrm{dy}\:=\frac{\mathrm{1}}{\mathrm{y}}\:+\mathrm{2log}\mid\frac{\mathrm{y}−\mathrm{1}}{\mathrm{y}+\mathrm{1}}\mid\:+\mathrm{C} \\ $$$$=\mathrm{e}^{−\mathrm{t}} \:+\mathrm{2log}\mid\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{1}}{\mathrm{e}^{\mathrm{t}} \:+\mathrm{1}}\mid\:+\mathrm{C}\:\mathrm{but}\:\mathrm{t}=\mathrm{argsh}\left(\frac{\mathrm{3x}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\mathrm{log}\left(\frac{\mathrm{3x}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{1}+\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}}\right)\:\Rightarrow \\ $$$$\sqrt{\mathrm{2}}\Psi=\frac{\mathrm{3x}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{1}+\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}}+\mathrm{2log}\mid\frac{\frac{\mathrm{3x}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{1}+\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}}−\mathrm{1}}{\frac{\mathrm{3x}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{1}+\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}}+\mathrm{1}}\mid\:+\mathrm{C}\:\Rightarrow \\ $$$$\Psi\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3x}+\sqrt{\mathrm{2}+\mathrm{9x}^{\mathrm{2}} }\right)\:+\sqrt{\mathrm{2}}\mathrm{log}\mid\frac{\mathrm{3x}+\sqrt{\mathrm{2}+\mathrm{9x}^{\mathrm{2}} }−\sqrt{\mathrm{2}}}{\mathrm{3x}+\sqrt{\mathrm{2}+\mathrm{9x}^{\mathrm{2}} }+\sqrt{\mathrm{2}}}\mid\:+\mathrm{C} \\ $$

Pin

Leave a Reply

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