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 85355 by TawaTawa1 last updated on 21/Mar/20

Commented by mathmax by abdo last updated on 21/Mar/20

I =∫ arctan((√(1+(√x))))dx changement (√(1+(√x)))=t give 1+(√x)=t^2 ⇒(√x)=t^2 −1 ⇒x =(t^2 −1)^2 ⇒dx =2(2t)(t^2 −1) ⇒ I =∫ arctan(t)(4t(t^2 −1))dt =4 ∫ (t^3 −t) arctant dt =_(byparts) 4{ (t^4 /4)−(t^2 /2) −∫ ((t^4 /4)−(t^2 /2))(dt/(1+t^2 ))} =t^4 −2t^2 −∫ ((t^4 −t^2 )/(t^2 +1))dt we have ∫ ((t^4 −t^2 )/(t^2 +1))dt =∫ ((t^2 (t^2 +1−2))/(t^2 +1))dt =∫ t^2 dt −∫ ((2t^2 )/(t^2 +1))dt =(t^3 /3)−2 ∫ ((t^2 +1−1)/(t^2 +1))dt =(t^3 /3)−2t +2arctan(t) +c ⇒ I =((√(1+(√x))))^4 −2((√(1+(√x))))^2 −(1/3)((√(1+(√x))))^3 +2(√(1+(√x))) +2arctan((√(1+(√x)))) +c

$${I}\:=\int\:{arctan}\left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right){dx}\:\:{changement}\:\sqrt{\mathrm{1}+\sqrt{{x}}}={t}\:{give} \\ $$$$\mathrm{1}+\sqrt{{x}}={t}^{\mathrm{2}} \:\Rightarrow\sqrt{{x}}={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow{x}\:=\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow{dx}\:=\mathrm{2}\left(\mathrm{2}{t}\right)\left({t}^{\mathrm{2}} −\mathrm{1}\right)\:\Rightarrow \\ $$$${I}\:=\int\:{arctan}\left({t}\right)\left(\mathrm{4}{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)\right){dt} \\ $$$$=\mathrm{4}\:\int\:\:\left({t}^{\mathrm{3}} −{t}\right)\:{arctant}\:{dt}\:\:=_{{byparts}} \:\:\mathrm{4}\left\{\:\:\frac{{t}^{\mathrm{4}} }{\mathrm{4}}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:−\int\:\:\left(\frac{{t}^{\mathrm{4}} }{\mathrm{4}}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right)\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\right\} \\ $$$$={t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \:−\int\:\:\frac{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:\:{we}\:{have} \\ $$$$\int\:\:\frac{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:=\int\:\:\frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:=\int\:{t}^{\mathrm{2}} \:{dt}\:−\int\:\frac{\mathrm{2}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$$=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}\:\int\:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}{t}\:\:+\mathrm{2}{arctan}\left({t}\right)\:+{c}\:\Rightarrow \\ $$$${I}\:=\left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)^{\mathrm{4}} −\mathrm{2}\left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)^{\mathrm{3}} \:+\mathrm{2}\sqrt{\mathrm{1}+\sqrt{{x}}}\:+\mathrm{2}{arctan}\left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)\:+{c} \\ $$

Commented by mathmax by abdo last updated on 21/Mar/20

error at line 5 I =t^4 −2t^2 −∫ ((t^4 −2t^2 )/(t^2 +1))dt =....

$${error}\:{at}\:{line}\:\mathrm{5}\:\:{I}\:={t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} −\int\:\frac{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:=.... \\ $$

Commented by TawaTawa1 last updated on 22/Mar/20

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by abdomathmax last updated on 23/Mar/20

you are welcome miss.

$${you}\:{are}\:{welcome}\:{miss}. \\ $$

Answered by mind is power last updated on 21/Mar/20

by part ∫tan^− ((√(1+(√x))))dx=(x−4)tan^− ((√(1+(√x))))−∫(x−4).(1/(4(√x).(√(1+(√x))) )).(dx/((2+(√x)))) =(x−4)tan^− ((√(1+(√x))))−∫(((√x)−2)/(4(√x).(√(1+(√x)))))dx u=(√x)⇒du=(dx/(2(√x))) ⇒(x−4)tan^− ((√(1+(√x))))−∫((u−2)/(2(√(1+u))))du =(x−4)tan^− ((√(1+(√x))))−∫((√(u+1))/2)du+∫((3du)/(2(√(1+u)))) =(x−4)tan^− ((√(1+(√x))))−(1/3)(u+1)^(3/2) +3(√(1+u))+c =(x−4)tan^− ((√(1+(√x))))+(((−(u+1))/3)+3)(√(1+u))+c u=(√x) (x−4)tan^− ((√(1+(√x))))+(−((√x)/3)+(8/3))(√(1+(√x)))+c =(x−4)tan^− ((√(1+(√x))))−((((√x)−8)/3))(√(1+(√x)))+c

$${by}\:{part} \\ $$$$\int{tan}^{−} \left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right){dx}=\left({x}−\mathrm{4}\right){tan}^{−} \left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)−\int\left({x}−\mathrm{4}\right).\frac{\mathrm{1}}{\mathrm{4}\sqrt{{x}}.\sqrt{\mathrm{1}+\sqrt{{x}}}\:\:}.\frac{{dx}}{\left(\mathrm{2}+\sqrt{{x}}\right)} \\ $$$$=\left({x}−\mathrm{4}\right){tan}^{−} \left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)−\int\frac{\sqrt{{x}}−\mathrm{2}}{\mathrm{4}\sqrt{{x}}.\sqrt{\mathrm{1}+\sqrt{{x}}}}{dx} \\ $$$${u}=\sqrt{{x}}\Rightarrow{du}=\frac{{dx}}{\mathrm{2}\sqrt{{x}}} \\ $$$$\Rightarrow\left({x}−\mathrm{4}\right){tan}^{−} \left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)−\int\frac{{u}−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{1}+{u}}}{du} \\ $$$$=\left({x}−\mathrm{4}\right){tan}^{−} \left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)−\int\frac{\sqrt{{u}+\mathrm{1}}}{\mathrm{2}}{du}+\int\frac{\mathrm{3}{du}}{\mathrm{2}\sqrt{\mathrm{1}+{u}}} \\ $$$$=\left({x}−\mathrm{4}\right){tan}^{−} \left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left({u}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{3}\sqrt{\mathrm{1}+{u}}+{c} \\ $$$$=\left({x}−\mathrm{4}\right){tan}^{−} \left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)+\left(\frac{−\left({u}+\mathrm{1}\right)}{\mathrm{3}}+\mathrm{3}\right)\sqrt{\mathrm{1}+{u}}+{c} \\ $$$${u}=\sqrt{{x}} \\ $$$$\left({x}−\mathrm{4}\right){tan}^{−} \left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)+\left(−\frac{\sqrt{{x}}}{\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{3}}\right)\sqrt{\mathrm{1}+\sqrt{{x}}}+{c} \\ $$$$=\left({x}−\mathrm{4}\right){tan}^{−} \left(\sqrt{\mathrm{1}+\sqrt{{x}}}\right)−\left(\frac{\sqrt{{x}}−\mathrm{8}}{\mathrm{3}}\right)\sqrt{\mathrm{1}+\sqrt{{x}}}+{c} \\ $$

Commented by TawaTawa1 last updated on 21/Mar/20

God bless you sir, i appreciate your time.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}. \\ $$

Commented by mind is power last updated on 23/Mar/20

withe Pleasur miss

$${withe}\:{Pleasur}\:{miss} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com