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Question Number 85166 by mathmax by abdo last updated on 19/Mar/20

find ∫ (x^2 −1)(√(x^2 +1))dx

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

Commented by john santu last updated on 21/Mar/20

let K = ∫ x^2 (√(x^2 +1)) dx K = (1/2)∫ x { 2x(√(x^2 +1)) } dx K= (1/2) ∫ x { (√(x^2 +1)) d(x^2 +1) } = (1/2){ (2/3) x (x^2 +1)^(3/2) −∫ (x^2 +1)^(3/2) dx} = (1/3)x(x^2 +1)^(3/2) −(1/2) ∫ sec^5 u du [ x = tan u ] H= ∫ sec^5 u du = ∫ sec^3 u d(tan u) H = sec^3 u tan u −∫ tan u (3sec^3 u tan u )du H= sec^3 u tan u −3∫(sec^5 u−sec^3 u)du 4H = sec^3 u tan u +3∫ sec^3 u du H = (1/4)x (x^2 +1)^(3/2) +(3/4)sec^3 u du let J = ∫ (√(x^2 +1)) dx , [ x=tan t ] J = ∫ sec^3 t dt = ∫ sec t d(tan t) J = sec t tan t − ∫ tan^2 t sec t dt J = sec t tan t − ∫ (sec^3 t−sec t) dt 2J = sec t tan t +ln ∣ x+(√(x^2 +1)) ∣ J= (1/2)sec t tan t + (1/2)ln ∣x+(√(x^2 +1)) ∣ ∴ H = (1/4)x(x^2 +1)^(3/2) +(7/8)x(x^2 +1)^(3/2) + (7/8)ln ∣x+(√(x^2 +1)) ∣ +c K= (1/3)x(x^2 +1)^(3/2) −(1/5)H ∴ K =(1/3)x(x^2 +1)^(3/2) −(9/(40))x(x^2 +1)^(3/2) − (7/(40)) ln ∣x+(√(x^2 +1)) ∣ +c K = ((13)/(120))x(x^2 +1)^(3/2) −(7/(40))ln ∣x+(√(x^2 +1)) ∣ + c

$$\mathrm{let}\:\mathrm{K}\:=\:\int\:\mathrm{x}^{\mathrm{2}} \:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx} \\ $$$$\mathrm{K}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{x}\:\left\{\:\mathrm{2x}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\right\}\:\mathrm{dx} \\ $$$$\mathrm{K}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\mathrm{x}\:\left\{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\:\right\} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{x}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\int\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{dx}\right\} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\mathrm{sec}\:^{\mathrm{5}} \mathrm{u}\:\mathrm{du} \\ $$$$\left[\:\mathrm{x}\:=\:\mathrm{tan}\:\mathrm{u}\:\right] \\ $$$$\mathrm{H}=\:\int\:\mathrm{sec}\:^{\mathrm{5}} \mathrm{u}\:\mathrm{du}\:=\:\int\:\mathrm{sec}\:^{\mathrm{3}} \mathrm{u}\:\mathrm{d}\left(\mathrm{tan}\:\mathrm{u}\right) \\ $$$$\mathrm{H}\:=\:\mathrm{sec}\:^{\mathrm{3}} \mathrm{u}\:\mathrm{tan}\:\mathrm{u}\:−\int\:\mathrm{tan}\:\mathrm{u}\:\left(\mathrm{3sec}\:^{\mathrm{3}} \mathrm{u}\:\mathrm{tan}\:\mathrm{u}\:\right)\mathrm{du} \\ $$$$\mathrm{H}=\:\mathrm{sec}\:^{\mathrm{3}} \mathrm{u}\:\mathrm{tan}\:\mathrm{u}\:−\mathrm{3}\int\left(\mathrm{sec}\:^{\mathrm{5}} \mathrm{u}−\mathrm{sec}\:^{\mathrm{3}} \mathrm{u}\right)\mathrm{du} \\ $$$$\mathrm{4H}\:=\:\mathrm{sec}\:^{\mathrm{3}} \mathrm{u}\:\mathrm{tan}\:\mathrm{u}\:+\mathrm{3}\int\:\mathrm{sec}\:^{\mathrm{3}} \mathrm{u}\:\mathrm{du} \\ $$$$\mathrm{H}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sec}\:^{\mathrm{3}} \mathrm{u}\:\mathrm{du} \\ $$$$\mathrm{let}\:\mathrm{J}\:=\:\int\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx}\:,\:\left[\:\mathrm{x}=\mathrm{tan}\:\mathrm{t}\:\right] \\ $$$$\mathrm{J}\:=\:\int\:\mathrm{sec}\:^{\mathrm{3}} \mathrm{t}\:\mathrm{dt}\:=\:\int\:\mathrm{sec}\:\mathrm{t}\:\mathrm{d}\left(\mathrm{tan}\:\mathrm{t}\right) \\ $$$$\mathrm{J}\:=\:\mathrm{sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:−\:\int\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{t}\:\mathrm{sec}\:\mathrm{t}\:\mathrm{dt} \\ $$$$\mathrm{J}\:=\:\mathrm{sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:−\:\int\:\left(\mathrm{sec}\:^{\mathrm{3}} \mathrm{t}−\mathrm{sec}\:\mathrm{t}\right)\:\mathrm{dt} \\ $$$$\mathrm{2J}\:=\:\mathrm{sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:+\mathrm{ln}\:\mid\:\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mid \\ $$$$\mathrm{J}=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mid \\ $$$$\therefore\:\mathrm{H}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{7}}{\mathrm{8}}\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} + \\ $$$$\frac{\mathrm{7}}{\mathrm{8}}\mathrm{ln}\:\mid\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mid\:+\mathrm{c} \\ $$$$\mathrm{K}=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{5}}\mathrm{H} \\ $$$$\therefore\:\mathrm{K}\:=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{9}}{\mathrm{40}}\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:− \\ $$$$\frac{\mathrm{7}}{\mathrm{40}}\:\mathrm{ln}\:\mid\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mid\:+\mathrm{c} \\ $$$$\mathrm{K}\:=\:\frac{\mathrm{13}}{\mathrm{120}}\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:−\frac{\mathrm{7}}{\mathrm{40}}\mathrm{ln}\:\mid\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mid\:+\:\mathrm{c} \\ $$

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

thank you sir john

$${thank}\:{you}\:{sir}\:{john} \\ $$

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

A =∫(x^2 −1)(√(x^2 +1))dx for this kind of integrals we do the changement x =sh(t) ⇒A =∫ (sh^2 t−1)ch(t)ch(t)dt A =∫ (sh^2 t−1)ch^2 t dt =∫ (((ch(2t)−1)/2)−1)(((ch(2t)+1)/2))dt =(1/4)∫ (ch(2t)−3)(ch(2t)+1)dt =(1/4)∫(ch^2 (2t)+ch(2t)−3ch(2t) −3)dt =(1/4)∫(ch^2 (2t)−2ch(2t)−3)dt =(1/8)∫ (1+ch(4t))dt −(1/2)∫ ch(2t)−(3/4)t +C =(1/8)t +(1/(32))sh(4t) −(1/4)sh(2t)−(3/4)t +C =−(5/8)t +(1/(16))sh(2t)ch(2t) −(1/2)sh(2t)ch(2t) +C =−(5/8)t +(1/8)sh(t)ch(t)(1+2sh^2 t) −sh(t)ch^2 t +C =−(5/8)ln(x+(√(1+x^2 ))) +(1/8)x(√(1+x^2 ))(1+2 x^2 ) −x (1+x^2 ) +C

$${A}\:=\int\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\:\:{for}\:{this}\:{kind}\:{of}\:{integrals}\:{we}\:{do}\:{the} \\ $$$${changement}\:{x}\:={sh}\left({t}\right)\:\Rightarrow{A}\:=\int\:\left({sh}^{\mathrm{2}} {t}−\mathrm{1}\right){ch}\left({t}\right){ch}\left({t}\right){dt} \\ $$$${A}\:=\int\:\left({sh}^{\mathrm{2}} {t}−\mathrm{1}\right){ch}^{\mathrm{2}} {t}\:{dt}\:=\int\:\left(\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)\left(\frac{{ch}\left(\mathrm{2}{t}\right)+\mathrm{1}}{\mathrm{2}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\:\left({ch}\left(\mathrm{2}{t}\right)−\mathrm{3}\right)\left({ch}\left(\mathrm{2}{t}\right)+\mathrm{1}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left({ch}^{\mathrm{2}} \left(\mathrm{2}{t}\right)+{ch}\left(\mathrm{2}{t}\right)−\mathrm{3}{ch}\left(\mathrm{2}{t}\right)\:−\mathrm{3}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left({ch}^{\mathrm{2}} \left(\mathrm{2}{t}\right)−\mathrm{2}{ch}\left(\mathrm{2}{t}\right)−\mathrm{3}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\:\left(\mathrm{1}+{ch}\left(\mathrm{4}{t}\right)\right){dt}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:{ch}\left(\mathrm{2}{t}\right)−\frac{\mathrm{3}}{\mathrm{4}}{t}\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}{t}\:\:+\frac{\mathrm{1}}{\mathrm{32}}{sh}\left(\mathrm{4}{t}\right)\:−\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)−\frac{\mathrm{3}}{\mathrm{4}}{t}\:+{C} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{8}}{t}\:+\frac{\mathrm{1}}{\mathrm{16}}{sh}\left(\mathrm{2}{t}\right){ch}\left(\mathrm{2}{t}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}{sh}\left(\mathrm{2}{t}\right){ch}\left(\mathrm{2}{t}\right)\:+{C} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{8}}{t}\:+\frac{\mathrm{1}}{\mathrm{8}}{sh}\left({t}\right){ch}\left({t}\right)\left(\mathrm{1}+\mathrm{2}{sh}^{\mathrm{2}} {t}\right)\:−{sh}\left({t}\right){ch}^{\mathrm{2}} {t}\:\:+{C} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{8}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+\frac{\mathrm{1}}{\mathrm{8}}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{2}\:{x}^{\mathrm{2}} \right)\:−{x}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \:\right)\:+{C} \\ $$

Answered by MJS last updated on 20/Mar/20

∫(x^2 −1)(√(x^2 +1))dx= [t=x+(√(x^2 +1)) → dx=((√(x^2 +1))/(x+(√(x^2 +1))))dt] =∫((t^3 /(16))−(t/4)−(5/(8t))−(1/(4t^3 ))+(1/(16t^5 )))dt= =(t^4 /(64))−(t^2 /8)−(5/8)ln t +(1/(8t^2 ))−(1/(64t^4 ))= =((t^8 −8t^6 +8t^2 −1)/(64t^4 ))−(5/8)ln t = =((x(2x^2 −3)(√(x^2 +1)))/8)−(5/8)ln (x+(√(x^2 +1))) +C

$$\int\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dt}\right] \\ $$$$=\int\left(\frac{{t}^{\mathrm{3}} }{\mathrm{16}}−\frac{{t}}{\mathrm{4}}−\frac{\mathrm{5}}{\mathrm{8}{t}}−\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{16}{t}^{\mathrm{5}} }\right){dt}= \\ $$$$=\frac{{t}^{\mathrm{4}} }{\mathrm{64}}−\frac{{t}^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{5}}{\mathrm{8}}\mathrm{ln}\:{t}\:+\frac{\mathrm{1}}{\mathrm{8}{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{64}{t}^{\mathrm{4}} }= \\ $$$$=\frac{{t}^{\mathrm{8}} −\mathrm{8}{t}^{\mathrm{6}} +\mathrm{8}{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{64}{t}^{\mathrm{4}} }−\frac{\mathrm{5}}{\mathrm{8}}\mathrm{ln}\:{t}\:= \\ $$$$=\frac{{x}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{8}}−\frac{\mathrm{5}}{\mathrm{8}}\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\:+{C} \\ $$

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

thank you sir mjs

$${thank}\:{you}\:{sir}\:{mjs} \\ $$

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