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Question Number 189189 by MikeH last updated on 13/Mar/23

Answered by cortano12 last updated on 13/Mar/23

$$\mathrm{it}\mathrm{should}\mathrm{be}\underset{-1/2}{\stackrel{1/2}{\int }}\sqrt{{\mathrm{x}}^2+1+\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^2+1}}\mathrm{dx}$$$$=\underset{-1/2}{\stackrel{1/2}{\int }}\sqrt{\frac{{\mathrm{x}}^2+\mathrm{x}+1}{2}}+\sqrt{\frac{{\mathrm{x}}^2-\mathrm{x}+1}{2}}\mathrm{dx}$$$$=\frac{1}{\sqrt{2}}\underset{-1/2}{\stackrel{1/2}{\int }}\sqrt{{(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4}}+\sqrt{{(\mathrm{x}-\frac{1}{2})}^2+\frac{3}{4}}\mathrm{dx}$$$$=\frac{1}{\sqrt{2}}{\underset{0}{\int }}^{\sinh {}^{-1}\left(\frac{2}{\sqrt{3}}\right)}\frac{\sqrt{3}}{2}\cosh {}^2\left(\frac{\sqrt{3}}{2}\theta \right)\mathrm{d}\theta$$$$+\frac{1}{\sqrt{2}}\underset{\sinh {}^{-1}\left(\frac{2}{\sqrt{3}}\right)}{\stackrel{0}{\int }}\frac{\sqrt{3}}{2}\cosh {}^2\left(\frac{\sqrt{3}}{2}\theta \right)\mathrm{d}\theta$$$$=\frac{\sqrt{7}}{2\sqrt{2}}+\frac{3}{4\sqrt{2}}\ln \left(\frac{2+\sqrt{7}}{\sqrt{3}}\right)$$

Commented by Frix last updated on 13/Mar/23

$$\mathrm{Ok}.\mathrm{But}:$$$$\mathrm{How}\mathrm{do}\mathrm{you}\mathrm{know}\mathrm{what}‘‘\mathrm{it}\mathrm{should}{\mathrm{be}}^{″}?$$

Commented by cortano12 last updated on 13/Mar/23

$$\mathrm{what}\mathrm{the}\mathrm{meaning}$$$$\sqrt{{\mathrm{x}}^2+1(\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^2+1)}}?$$

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