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Question Number 146212 by puissant last updated on 12/Jul/21

K=∫(1/( (√(1+x^3 ))))dx

$$\mathrm{K}=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }}\mathrm{dx} \\ $$

Answered by Ar Brandon last updated on 12/Jul/21

I=∫(dx/( (√(1+x^3 ))))=∫(dx/( (√(1+(x^(3/2) )^2 )))), u=x^(3/2) ⇒du=(3/2)x^(1/2) dx=(3/2)u^(1/3) dx =(2/3)∫(u^(−1/3) /( (√(1+u^2 ))))du, u=sinhϑ⇒du=coshϑdϑ =(2/3)∫(((sinhϑ)^(−1/3) )/(coshϑ))∙coshϑdϑ=(2/3)∫(sinhϑ)^(−1/3) dϑ =x _2 F_1 ((1/3); (1/2); (4/3); −x^3 ) , Hypergeome^� trique. ∫_0 ^∞ (dx/( (√(1+x^3 )))), u=x^3 ⇒(1/3)u^(−(2/3)) du=dx =(1/3)∫_0 ^∞ (u^(−(2/3)) /( (√(1+u))))du=(1/3)β((1/3), (1/6))=((Γ((1/3))Γ((1/6)))/(3Γ((1/2))))=((Γ((1/3))Γ((1/6)))/(3(√π))) ≈2.80436421

$$\mathrm{I}=\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }}=\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\left(\mathrm{x}^{\mathrm{3}/\mathrm{2}} \right)^{\mathrm{2}} }},\:\mathrm{u}=\mathrm{x}^{\mathrm{3}/\mathrm{2}} \Rightarrow\mathrm{du}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}^{\mathrm{1}/\mathrm{2}} \mathrm{dx}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{u}^{\mathrm{1}/\mathrm{3}} \mathrm{dx} \\ $$$$\:\:=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{u}^{−\mathrm{1}/\mathrm{3}} }{\:\sqrt{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }}\mathrm{du},\:\mathrm{u}=\mathrm{sinh}\vartheta\Rightarrow\mathrm{du}=\mathrm{cosh}\vartheta\mathrm{d}\vartheta \\ $$$$\:\:=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\left(\mathrm{sinh}\vartheta\right)^{−\mathrm{1}/\mathrm{3}} }{\mathrm{cosh}\vartheta}\centerdot\mathrm{cosh}\vartheta\mathrm{d}\vartheta=\frac{\mathrm{2}}{\mathrm{3}}\int\left(\mathrm{sinh}\vartheta\right)^{−\mathrm{1}/\mathrm{3}} \mathrm{d}\vartheta \\ $$$$\:\:=\mathrm{x}\underset{\mathrm{2}} {\:}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{4}}{\mathrm{3}};\:−\mathrm{x}^{\mathrm{3}} \right)\:,\:\mathrm{Hypergeom}\acute {\mathrm{e}trique}. \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }},\:\mathrm{u}=\mathrm{x}^{\mathrm{3}} \Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\mathrm{u}^{−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{du}=\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\frac{\mathrm{2}}{\mathrm{3}}} }{\:\sqrt{\mathrm{1}+\mathrm{u}}}\mathrm{du}=\frac{\mathrm{1}}{\mathrm{3}}\beta\left(\frac{\mathrm{1}}{\mathrm{3}},\:\frac{\mathrm{1}}{\mathrm{6}}\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{3}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{3}\sqrt{\pi}} \\ $$$$\approx\mathrm{2}.\mathrm{80436421} \\ $$

Commented by puissant last updated on 12/Jul/21

merci mais on ne peut faire ca avec le theoreme des residus..??

$$\mathrm{merci}\:\mathrm{mais}\:\mathrm{on}\:\mathrm{ne}\:\:\mathrm{peut}\:\mathrm{faire}\:\mathrm{ca}\:\mathrm{avec}\:\mathrm{le}\:\mathrm{theoreme} \\ $$$$\mathrm{des}\:\mathrm{residus}..?? \\ $$

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