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Question Number 121519 by benjo_mathlover last updated on 09/Nov/20

$$\int \frac{1-{\mathrm{x}}^4}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^2+1}}\mathrm{dx}?$$

Answered by liberty last updated on 09/Nov/20

$$\int \frac{1-{\mathrm{x}}^4}{{\mathrm{x}}^3\sqrt{{\mathrm{x}}^2+1+{\mathrm{x}}^{-2}}}\mathrm{dx}=$$$$\int \frac{{\mathrm{x}}^{-3}-\mathrm{x}}{\sqrt{{\mathrm{x}}^2+{\mathrm{x}}^{-2}+1}}\mathrm{dx}$$$$\mathrm{let}\mathrm{v}={\mathrm{x}}^2+{\mathrm{x}}^{-2}+1\to \mathrm{dv}=2\mathrm{x}-{2\mathrm{x}}^{-3}\mathrm{dx}$$$$\mathrm{dv}=-2({\mathrm{x}}^{-3}-\mathrm{x})\mathrm{dx}$$$$$$$$\int \frac{-\frac{1}{2}\mathrm{dv}}{\sqrt{\mathrm{v}}}=-\frac{1}{2}\int {\mathrm{v}}^{-\frac{1}{2}}\mathrm{dv}$$$$=-\sqrt{\mathrm{v}}+\mathrm{c}=-\sqrt{{\mathrm{x}}^2+{\mathrm{x}}^{-2}+1}+\mathrm{c}$$$$=-\frac{\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^2+1}}{\mathrm{x}}+\mathrm{c}$$

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