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Question Number 152631 by mnjuly1970 last updated on 30/Aug/21

Answered by Olaf_Thorendsen last updated on 30/Aug/21

$$\mathrm{I}={\int }_0^1\frac{\ln (1-x^2+x^4)}{x^2}dx$$$$\mathrm{I}={[-\frac{\ln (1-x^2+x^4)}{x}]}_0^1-{\int }_0^1-\frac{1}{x}\left(\frac{4x^3-2x}{x^4-x^2+1}\right)dx$$$$\mathrm{I}=2{\int }_0^1\frac{2x^2-1}{x^4-x^2+1}dx$$$$\mathrm{I}=2{\int }_0^1\frac{2x^2-1}{(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)}dx$$$$etc...$$

Commented by SANOGO last updated on 31/Aug/21

$$good$$

Commented by mnjuly1970 last updated on 31/Aug/21

$$thanksalot...$$

Commented by Ar Brandon last updated on 02/Sep/21

$$I=2{\int }_0^1\frac{2x^2-1}{x^4-x^2+1}dx=$$$$2x^2-1=x^2-1+\frac{1}{2}[(x^2-1)+(x^2+1)]$$$$=\frac{3}{2}(x^2-1)+\frac{1}{2}(x^2+1)$$$$I=3{\int }_0^1\frac{x^2-1}{x^4-x^2+1}dx+{\int }_0^1\frac{x^2+1}{x^4-x^2+1}dx$$$$=3{\int }_0^1\frac{1-\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}dx+{\int }_0^1\frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}dx$$$$=3{\int }_0^1\frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-3}dx+{\int }_0^1\frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+1}dx$$$$=-\sqrt{3}{\left[\mathrm{argth}\left(\frac{x^2+1}{\sqrt{3}x}\right)\right]}_0^1+{\left[\arctan \left(\frac{x^2-1}{\sqrt{3}x}\right)\right]}_0^1$$$$=\frac{\sqrt{3}}{2}{[\ln \mid \frac{x^2-\sqrt{3}x+1}{x^2+\sqrt{3}x+1}\mid ]}_0^1+\frac{\pi }{2}=\frac{\sqrt{3}}{2}\ln \left(\frac{2-\sqrt{3}}{2+\sqrt{3}}\right)+\frac{\pi }{2}$$$$=\frac{\sqrt{3}}{2}\ln \left(\frac{1}{{(2+\sqrt{3})}^2}\right)+\frac{\pi }{2}=\frac{\pi }{2}-\sqrt{3}\ln (2+\sqrt{3})$$$$=\frac{1}{2}(\pi -2\sqrt{3}\ln (2+\sqrt{3}))$$

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