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Question Number 192712 by cortano12 last updated on 25/May/23

$$\underset{-1/2}{\stackrel{1/2}{\int }}\sqrt{{\mathrm{x}}^2+1+\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^2+1}}\mathrm{dx}=?$$

Answered by horsebrand11 last updated on 25/May/23

$$I=2\underset{0}{\stackrel{1/2}{\int }}\sqrt{x^2+1+\sqrt{(x^2+x+1)(x^2-x+1)}}dx$$$$=\sqrt{2}\underset{0}{\stackrel{1/2}{\int }}\sqrt{{\left(\sqrt{x^2+x+1}\right)+\sqrt{x^2-x+1})}^2}dx$$$$=\sqrt{2}\underset{0}{\stackrel{1/2}{\int }}(\sqrt{x^2+x+1}+\sqrt{x^2-x+1})dx$$$$=\frac{1}{\sqrt{2}}(\underset{0}{\stackrel{1/2}{\int }}\sqrt{{(2x+1)}^2+3}dx+\underset{-1/2}{\stackrel{0}{\int }}\sqrt{{(2x+1)}^2+3}dx)$$$$=\frac{1}{\sqrt{2}}\underset{-1/2}{\stackrel{1/2}{\int }}\sqrt{{(2x+1)}^2+3}dx$$$$let2x+1=\sqrt{3}\tan \alpha$$$$I=\frac{3}{2\sqrt{2}}\underset{0}{\stackrel{\arctan (2/\sqrt{3)}}{\int }}\sec {}^3\alpha d\alpha$$$$=\frac{3}{4\sqrt{2}}{(\sec \alpha \tan \alpha +\ln (\sec \alpha +\tan \alpha )]}_0^{\arctan \left(\frac{2}{\sqrt{3}}\right)}$$$$=2\sqrt{7}+3\ln \left(\frac{2+\sqrt{7}}{\sqrt{3}}\right)$$

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