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Question Number 145378 by mnjuly1970 last updated on 04/Jul/21

∫_0 ^( ∞) ((√( 1+ x^4 )) −x^( 2) )dx=?

$$ \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\left(\sqrt{\:\mathrm{1}+\:\mathrm{x}^{\mathrm{4}} }\:−\mathrm{x}^{\:\mathrm{2}} \:\right)\mathrm{dx}=? \\ $$

Answered by qaz last updated on 05/Jul/21

∫_0 ^∞ ((√(1+x^4 ))−x^2 )dx =x((√(1+x^4 ))−x^2 )∣_0 ^∞ −2∫_0 ^∞ (((1+x^4 −1)/( (√(1+x^4 ))))−x^2 )dx =(x/( (√(1+x^4 ))+x^2 ))∣_0 ^∞ −2∫_0 ^∞ ((√(1+x^4 ))−x^2 )dx+2∫_0 ^∞ (dx/( (√(1+x^4 )))) =(2/3)∫_0 ^∞ (dx/( (√(1+x^4 )))) =(2/3)∙(1/4)∫_0 ^∞ (u^(−3/4) /((1+u)^(1/2) ))du............x^4 →u =(1/6)B((1/4),(1/4)) =((Γ^2 ((1/4)))/(6(√π)))

$$\int_{\mathrm{0}} ^{\infty} \left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }−\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$=\mathrm{x}\left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }−\mathrm{x}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\infty} −\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}+\mathrm{x}^{\mathrm{4}} −\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$=\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }+\mathrm{x}^{\mathrm{2}} }\mid_{\mathrm{0}} ^{\infty} −\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left(\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }−\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}+\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\mathrm{3}/\mathrm{4}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{1}/\mathrm{2}} }\mathrm{du}............\mathrm{x}^{\mathrm{4}} \rightarrow\mathrm{u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{B}\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{6}\sqrt{\pi}} \\ $$

Commented by mnjuly1970 last updated on 05/Jul/21

very nice mr qaz thx alot...

$$\:\:{very}\:{nice}\:{mr}\:{qaz} \\ $$$${thx}\:{alot}... \\ $$

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