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Question Number 65355 by mathmax by abdo last updated on 28/Jul/19

find ∫ (dx/(√((x+1)(x+2)(x+3))))

$${find}\:\int\:\:\:\frac{{dx}}{\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)}} \\ $$

Commented by Prithwish sen last updated on 29/Jul/19

∫(dx/(√({(x+2)^2 −1}(x+2)))) putting (x+2)=a =∫(da/(√(a(a^2 −1)))) again putting a=secθ = ∫(√(secθ)) dθ ????

$$\int\frac{\mathrm{dx}}{\sqrt{\left\{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}\right\}\left(\mathrm{x}+\mathrm{2}\right)}}\:\:\:\mathrm{putting}\:\left(\mathrm{x}+\mathrm{2}\right)=\mathrm{a} \\ $$$$\:=\int\frac{\mathrm{da}}{\sqrt{\mathrm{a}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{1}\right)}}\:\:\mathrm{again}\:\mathrm{putting}\:\mathrm{a}=\mathrm{sec}\theta \\ $$$$=\:\int\sqrt{\mathrm{sec}\theta}\:\mathrm{d}\theta\:\:???? \\ $$

Answered by ajfour last updated on 29/Jul/19

x+2=t ∫(dx/(√(t(t^2 −1)))) let t=sec^2 θ ⇒ dt=2sec^2 θtan θdθ ∫((2sec θtan θdθ)/(√(sec^4 θ−1))) let sec θ=z ∫((2dz)/(√(z^4 −1)))= ?

$${x}+\mathrm{2}={t} \\ $$$$\int\frac{{dx}}{\sqrt{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}} \\ $$$${let}\:{t}=\mathrm{sec}\:^{\mathrm{2}} \theta\:\Rightarrow\:\:{dt}=\mathrm{2sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:\theta{d}\theta \\ $$$$\int\frac{\mathrm{2sec}\:\theta\mathrm{tan}\:\theta{d}\theta}{\sqrt{\mathrm{sec}\:^{\mathrm{4}} \theta−\mathrm{1}}} \\ $$$${let}\:\:\mathrm{sec}\:\theta={z} \\ $$$$\int\frac{\mathrm{2}{dz}}{\sqrt{{z}^{\mathrm{4}} −\mathrm{1}}}=\:? \\ $$

Answered by MJS last updated on 29/Jul/19

∫(dx/(√((x+1)(x+2)(x+3))))= [t=arccosh (x+2) → dx=dt(√((x+1)(x+3)))] =∫(dt/(√(cosh t)))= [u=(t/2)i → dt=−2idu] =−2i∫(du/(√(cos 2u)))=−2i∫(du/(√(1−2sin^2 u)))= this is an incomplete elliptic integral =−2iF(u∣2) ...

$$\int\frac{{dx}}{\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arccosh}\:\left({x}+\mathrm{2}\right)\:\rightarrow\:{dx}={dt}\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)}\right] \\ $$$$=\int\frac{{dt}}{\sqrt{\mathrm{cosh}\:{t}}}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{{t}}{\mathrm{2}}\mathrm{i}\:\rightarrow\:{dt}=−\mathrm{2i}{du}\right] \\ $$$$=−\mathrm{2i}\int\frac{{du}}{\sqrt{\mathrm{cos}\:\mathrm{2}{u}}}=−\mathrm{2i}\int\frac{{du}}{\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{u}}}= \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{an}\:\mathrm{incomplete}\:\mathrm{elliptic}\:\mathrm{integral} \\ $$$$=−\mathrm{2i}{F}\left({u}\mid\mathrm{2}\right) \\ $$$$... \\ $$

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