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Question Number 133004 by liberty last updated on 18/Feb/21

$$\mathrm{If}\int \frac{\tan \mathrm{x}}{1+\tan \mathrm{x}+\tan {}^2\mathrm{x}}\mathrm{dx}=\mathrm{x}-\frac{k}{\sqrt{A}}{\tan }^{-1}\left(\frac{k\tan x+1}{\sqrt{A}}\right)+\mathrm{C}$$$$\mathrm{where}\mathrm{C}\mathrm{is}\mathrm{constant}\mathrm{of}\mathrm{integration}.$$$$\mathrm{then}\mathrm{the}\mathrm{ordered}\mathrm{pair}(k,\mathrm{A})\mathrm{is}$$$$\mathrm{equal}\mathrm{to}$$

Answered by EDWIN88 last updated on 18/Feb/21

$$\mathrm{consider}\tan \mathrm{x}=1+\tan {}^2\mathrm{x}+\tan \mathrm{x}-\sec {}^2\mathrm{x}$$$$\mathrm{I}=\int \frac{1+\tan {}^2\mathrm{x}+\tan \mathrm{x}-\sec {}^2\mathrm{x}}{1+\tan \mathrm{x}+\tan {}^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{I}=\int 1-\frac{\sec {}^2\mathrm{x}}{1+\tan \mathrm{x}+\tan {}^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{I}=\mathrm{x}-\int \frac{\sec {}^2\mathrm{x}}{1+\tan \mathrm{x}+\tan {}^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{put}\mathrm{u}=\tan \mathrm{x}$$$$\mathrm{I}=\mathrm{x}-\int \frac{\mathrm{du}}{1+\mathrm{u}+{\mathrm{u}}^2}=\mathrm{x}-\int \frac{\mathrm{du}}{{(\mathrm{u}+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$\mathrm{I}=\mathrm{x}-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\mathrm{u}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+\mathrm{C}$$$$\mathrm{I}=\mathrm{x}-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan \mathrm{x}+1}{\sqrt{3}}\right)+\mathrm{C}$$$$\mathrm{we}\mathrm{find}\to \{\begin{array}{l}k=2\\ A=3\end{array}\Rightarrow (k,\mathrm{A})=(2,3)$$

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