Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 94143 by john santu last updated on 17/May/20

∫ (dx/((x+(√(1+x^2 )))^2 )) =

$$\int\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:= \\ $$

Commented by abdomathmax last updated on 17/May/20

I =∫  (dx/((x+(√(1+x^2 )))^2 )) we do the chsngement x =sh(t)  ⇒I =∫   ((cht)/((sht +cht)))dt  =∫   ((e^t  +e^(−t) )/(2(((e^t  +e^(−t) )/2)+((e^t −e^(−t) )/2))^2 ))dt  =(1/2) ∫ ((e^t  +e^(−t) )/e^(2t) )dt =(1/2)∫ e^(−2t) (e^t  +e^(−t) )dt  =(1/2) ∫(e^(−t)  +e^(−3t) )dt =−(1/2)e^(−t) −(1/6)e^(−3t)  +C  t=argsh(x)=ln(x+(√(1+x^2 ))) ⇒  I =−(1/(2(x+(√(1+x^2 ))))) −(1/(6(x+(√(1+x^2 )))^3 )) +C

$${I}\:=\int\:\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:{we}\:{do}\:{the}\:{chsngement}\:{x}\:={sh}\left({t}\right) \\ $$$$\Rightarrow{I}\:=\int\:\:\:\frac{{cht}}{\left({sht}\:+{cht}\right)}{dt}\:\:=\int\:\:\:\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}\left(\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}+\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}\right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{{e}^{{t}} \:+{e}^{−{t}} }{{e}^{\mathrm{2}{t}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:{e}^{−\mathrm{2}{t}} \left({e}^{{t}} \:+{e}^{−{t}} \right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\left({e}^{−{t}} \:+{e}^{−\mathrm{3}{t}} \right){dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{t}} −\frac{\mathrm{1}}{\mathrm{6}}{e}^{−\mathrm{3}{t}} \:+{C} \\ $$$${t}={argsh}\left({x}\right)={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{2}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}\:−\frac{\mathrm{1}}{\mathrm{6}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{3}} }\:+{C} \\ $$

Commented by i jagooll last updated on 17/May/20

what the meaning cht ? cosh  (t)?

$$\mathrm{what}\:\mathrm{the}\:\mathrm{meaning}\:\mathrm{cht}\:?\:\mathrm{cosh}\:\:\left(\mathrm{t}\right)? \\ $$

Commented by mathmax by abdo last updated on 17/May/20

cht =((e^t  +e^(−t) )/2) (→hyperbolic cos)  sh(t) =((e^t −e^(−t) )/2)(→hyperbolic sin)

$${cht}\:=\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}\:\left(\rightarrow{hyperbolic}\:{cos}\right) \\ $$$${sh}\left({t}\right)\:=\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}\left(\rightarrow{hyperbolic}\:{sin}\right) \\ $$

Answered by john santu last updated on 17/May/20

(1/(x+(√(1+x^2 )))) × ((x−(√(1+x^2 )))/(x−(√(1+x^2 )))) = ((x−(√(1+x^2 )))/(−1))  ∫ (dx/((x+(√(1+x^2 )))^2 )) = ∫ (((x−(√(1+x^2 )))/(−1)))^2 dx  =∫ x^2 −2x(√(1+x^2 ))+1+x^2  dx  = ∫ (2x^2 +1−2x(√(1+x^2 ))) dx   = (2/3)x^3 +x−(2/3) (√((1+x^2 )^3 )) + c

$$\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:×\:\frac{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:=\:\frac{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{−\mathrm{1}} \\ $$$$\int\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:=\:\int\:\left(\frac{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{−\mathrm{1}}\right)^{\mathrm{2}} {dx} \\ $$$$=\int\:{x}^{\mathrm{2}} −\mathrm{2}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{1}+{x}^{\mathrm{2}} \:{dx} \\ $$$$=\:\int\:\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:{dx}\: \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} +{x}−\frac{\mathrm{2}}{\mathrm{3}}\:\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:+\:{c}\: \\ $$

Commented by i jagooll last updated on 17/May/20

cool man ����

Terms of Service

Privacy Policy

Contact: info@tinkutara.com