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Question Number 125895 by john_santu last updated on 15/Dec/20

$$\int \frac{dx}{\tan {}^{4}x+1}?$$

Answered by Dwaipayan Shikari last updated on 15/Dec/20

$$\int \frac{dx}{{tan}^{4}x+1}tanx=t\Rightarrow {sec}^{2}x=\frac{dt}{dx}$$$$=\int \frac{dt}{(t^2+1)(t^4+1)}=\frac{1}{2}\int \frac{1}{(t^2+1)}-\frac{t^2-1}{t^4+1}=\frac{1}{2}{tan}^{-1}t-\frac{1}{2}\int \frac{t^2+1}{t^4+1}+\int \frac{1}{t^4+1}$$$$=\frac{x}{2}-\frac{1}{4}\int \frac{1}{(t^2-\sqrt{2}t+1)}+\frac{1}{(t^2+\sqrt{2}t+1)}+\int \frac{1}{(t^2-\sqrt{2}t+1)(t^2+\sqrt{2}t+1)}$$$$=\frac{x}{2}-\frac{1}{2\sqrt{2}}({tan}^{-1}(\sqrt{2}t-1)+{tan}^{-1}(\sqrt{2}t+1))-\frac{1}{2\sqrt{2}}\int \frac{t-\sqrt{2}}{t^2-\sqrt{2}t+1}-\frac{t+\sqrt{2}}{t^2+\sqrt{2}t+1}$$$$=\frac{x}{2}-\frac{1}{2\sqrt{2}}{tan}^{-1}\frac{\sqrt{2}tanx}{1-{tan}^{2}x}-\frac{1}{2\sqrt{2}}\int \frac{2t-\sqrt{2}}{t^2-\sqrt{2}t+1}-\frac{2t+\sqrt{2}}{t^2+\sqrt{2}t+1}-\frac{1}{2\sqrt{2}}\int \frac{1}{t^2+\sqrt{2}t+1}-\frac{1}{t^2-\sqrt{2}t+1}$$$$=\frac{x}{2}-\frac{1}{2\sqrt{2}}{tan}^{-1}\frac{\sqrt{2}tanx}{1-{tan}^{2}x}-\frac{1}{2\sqrt{2}}log\left(\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right)-\frac{1}{2}\left({tan}^{-1}\frac{1}{t^2}\right)+C$$$$=\frac{x}{2}-\frac{1}{2}(\frac{1}{\sqrt{2}}{tan}^{-1}\frac{\sqrt{2}tanx}{1-{tan}^{2}x}-{tan}^{-1}\frac{1}{{tan}^{2}x})-\frac{1}{2\sqrt{2}}log\left(\frac{{tan}^{2}x-\sqrt{2}tanx+1}{{tan}^{2}x+\sqrt{2}tanx+1}\right)+C$$$$$$

Commented by MJS_new last updated on 15/Dec/20

$$\mathrm{I}\mathrm{get}\left(\mathrm{just}\mathrm{more}\mathrm{transforming}\mathrm{I}\mathrm{guess}\right)$$$$\frac{x}{2}+\frac{\sqrt{2}}{8}(\ln \mid \sqrt{2}+\sin 2x\mid -\ln \mid \sqrt{2}-\sin 2x\mid )+C$$

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