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Question Number 108071 by mathdave last updated on 14/Aug/20

Answered by mathmax by abdo last updated on 14/Aug/20

$$\mathrm{sir}\mathrm{consider}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^1\frac{\ln (\mathrm{a}+{\mathrm{x}}^2)}{1+{\mathrm{x}}^4}\mathrm{dx}\mathrm{with}\mathrm{a}>0\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^1\frac{1}{({\mathrm{x}}^2+\mathrm{a})({\mathrm{x}}^4+1)}\mathrm{dx}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{({\mathrm{x}}^2+\mathrm{a})({\mathrm{x}}^4+1)}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{({\mathrm{x}}^2+\mathrm{a})({({\mathrm{x}}^2+1)}^2-{2\mathrm{x}}^2)}=\frac{1}{({\mathrm{x}}^2+\mathrm{a})({\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1)({\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1)}$$$$=\frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2+\mathrm{a}}+\frac{\mathrm{cx}+\mathrm{d}}{{\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1}+\frac{\mathrm{ex}+\mathrm{f}}{{\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1}\mathrm{calculste}\mathrm{the}\mathrm{coeffcient}$$$$\mathrm{explicite}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)\mathrm{then}\mathrm{f}\left(\mathrm{a}\right)\mathrm{and}\mathrm{take}\mathrm{a}=1\mathrm{you}\mathrm{arrive}\mathrm{the}\mathrm{value}$$$$\mathrm{but}\mathrm{you}\mathrm{find}\mathrm{a}\mathrm{lots}\mathrm{of}\mathrm{calculus}.....$$$$$$

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