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Question Number 144402 by rexford last updated on 25/Jun/21

Commented by justtry last updated on 25/Jun/21

Commented by rexford last updated on 27/Jun/21

thanks very much for your time I am very grateful

$${thanks}\:{very}\:{much}\:{for}\:{your}\:{time} \\ $$$${I}\:{am}\:{very}\:{grateful} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 25/Jun/21

Ψ=∫ (dx/(x^2 (√(27x^2 +6x−2)))) ⇒ Δ^′ =3^2 −27.(−2)=9+54 =63 ⇒x_1 =((−3+(√(63)))/(27)) x_2 =((−3−(√(63)))/(27)) ⇒Ψ=∫ (dx/(x^2 (√(27(x−x_1 )(x−x_2 ))))) =(1/( (√(27))))∫ (dx/(x^2 (√(x−x_1 ))(√(x−x_2 )))) =_((√(x−x_1 ))=t) (1/( (√(27))))∫ ((2tdt)/((t^2 +x_1 )^2 t(√(t^2 +x_1 −x_2 )))) =(2/( (√(27))))∫(dt/((t^2 +x_1 )^2 (√(t^2 +x_(1 ) −x_2 )))) we have x_1 −x_2 =((2(√(63)))/(27)) so we do the cha7gement t=(√((2(√(63)))/(27)))shy ⇒Ψ=(2/( (√(27))))∫ ((λ_0 chy dy)/((λ_0 ^2 sh^2 y +x_1 )^2 λ_0 chy)) =(2/( (√(27))))∫ (dy/((λ_0 ^2 sh^2 y +x_1 )^2 )) (λ_0 =(√((2(√(63)))/(27)))) ....be continued...

$$\Psi=\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{27x}^{\mathrm{2}} \:+\mathrm{6x}−\mathrm{2}}}\:\Rightarrow \\ $$$$\Delta^{'} \:=\mathrm{3}^{\mathrm{2}} −\mathrm{27}.\left(−\mathrm{2}\right)=\mathrm{9}+\mathrm{54}\:=\mathrm{63}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\frac{−\mathrm{3}+\sqrt{\mathrm{63}}}{\mathrm{27}} \\ $$$$\mathrm{x}_{\mathrm{2}} =\frac{−\mathrm{3}−\sqrt{\mathrm{63}}}{\mathrm{27}}\:\Rightarrow\Psi=\int\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{27}\left(\mathrm{x}−\mathrm{x}_{\mathrm{1}} \right)\left(\mathrm{x}−\mathrm{x}_{\mathrm{2}} \right)}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{27}}}\int\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}−\mathrm{x}_{\mathrm{1}} }\sqrt{\mathrm{x}−\mathrm{x}_{\mathrm{2}} }} \\ $$$$=_{\sqrt{\mathrm{x}−\mathrm{x}_{\mathrm{1}} }=\mathrm{t}} \:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{27}}}\int\:\:\:\frac{\mathrm{2tdt}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{2}} \mathrm{t}\sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{x}_{\mathrm{1}} −\mathrm{x}_{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{27}}}\int\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{2}} \sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{x}_{\mathrm{1}\:} −\mathrm{x}_{\mathrm{2}} }} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{x}_{\mathrm{1}} −\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{2}\sqrt{\mathrm{63}}}{\mathrm{27}}\:\:\mathrm{so}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{cha7gement}\:\mathrm{t}=\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{63}}}{\mathrm{27}}}\mathrm{shy} \\ $$$$\Rightarrow\Psi=\frac{\mathrm{2}}{\:\sqrt{\mathrm{27}}}\int\:\:\:\frac{\lambda_{\mathrm{0}} \mathrm{chy}\:\mathrm{dy}}{\left(\lambda_{\mathrm{0}} ^{\mathrm{2}} \:\mathrm{sh}^{\mathrm{2}} \mathrm{y}\:+\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{2}} \lambda_{\mathrm{0}} \mathrm{chy}} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{27}}}\int\:\:\frac{\mathrm{dy}}{\left(\lambda_{\mathrm{0}} ^{\mathrm{2}} \:\mathrm{sh}^{\mathrm{2}} \mathrm{y}\:+\mathrm{x}_{\mathrm{1}} \right)^{\mathrm{2}} }\:\:\:\:\left(\lambda_{\mathrm{0}} =\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{63}}}{\mathrm{27}}}\right) \\ $$$$....\mathrm{be}\:\mathrm{continued}... \\ $$

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