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Question Number 121384 by john santu last updated on 07/Nov/20

$$\int \sqrt{1+{\mathrm{x}}^4}\mathrm{dx}$$

Commented by MJS_new last updated on 07/Nov/20

$$\mathrm{not}\mathrm{possible}\mathrm{to}\mathrm{solve}\mathrm{using}\mathrm{only}\mathrm{elementary}$$$$\mathrm{calculus}.\mathrm{I}\mathrm{guess}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{substitute}\mathrm{and}$$$$\mathrm{transform}\mathrm{to}\mathrm{get}\mathrm{an}\mathrm{elliptic}\mathrm{integral}$$

Commented by peter frank last updated on 08/Nov/20

$$\mathrm{true}\mathrm{sir}....\mathrm{may}\mathrm{be}\int \mathrm{x}\sqrt{1+{\mathrm{x}}^4}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 07/Nov/20

$$\mathrm{A}=\int \sqrt{1+{\mathrm{x}}^4}\mathrm{dx}\mathrm{by}\mathrm{parts}$$$$\mathrm{A}=\mathrm{x}\sqrt{1+{\mathrm{x}}^4}-\int \mathrm{x}.\frac{{4\mathrm{x}}^3}{\sqrt{1+{\mathrm{x}}^4}}\mathrm{dx}=\mathrm{x}\sqrt{1+{\mathrm{x}}^4}-4\int \frac{1+{\mathrm{x}}^4-1}{\sqrt{1+{\mathrm{x}}^4}}\mathrm{dx}$$$$=\mathrm{x}\sqrt{1+{\mathrm{x}}^4}-4\int \sqrt{1+{\mathrm{x}}^4}+4\int \frac{\mathrm{dx}}{\sqrt{1+{\mathrm{x}}^4}}\Rightarrow$$$$5\mathrm{A}=\mathrm{x}\sqrt{1+{\mathrm{x}}^4}+4\int \frac{\mathrm{dx}}{\sqrt{1+{\mathrm{x}}^4}}\Rightarrow \mathrm{A}=\frac{\mathrm{x}}{5}\sqrt{1+{\mathrm{x}}^4}+\frac{4}{5}\int \frac{\mathrm{dx}}{\sqrt{1+{\mathrm{x}}^4}}$$$$\mathrm{changement}{\mathrm{x}}^2=\mathrm{tant}\mathrm{give}\mathrm{x}=\sqrt{\mathrm{tant}}\Rightarrow$$$$\int \frac{\mathrm{dx}}{\sqrt{1+{\mathrm{x}}^4}}=\int \frac{(1+{\tan }^2\mathrm{t})}{2\sqrt{\mathrm{tant}}\times \sqrt{1+{\tan }^2\mathrm{t}}}\mathrm{dt}=\frac{1}{2}\int \frac{\sqrt{1+{\tan }^2\mathrm{t}}}{\sqrt{\mathrm{tant}}}\mathrm{du}$$$$....\mathrm{be}\mathrm{continued}....$$

Answered by peter frank last updated on 07/Nov/20

$$\mathrm{let}{\mathrm{x}}^2=\tan \theta$$$$2\mathrm{xdx}=\sec {}^2\theta \mathrm{d}\theta$$$$\mathrm{dx}=\frac{\sec {}^2\theta \mathrm{d}\theta }{2\sqrt{\tan \theta }}$$$$\int \sqrt{1+{\mathrm{x}}^4}\mathrm{dx}$$$$\int \sqrt{1+\tan {}^2\theta }\mathrm{dx}$$$$\int \sqrt{1+\tan {}^2\theta }\frac{\sec {}^2\theta \mathrm{d}\theta }{2\sqrt{\tan \theta }}$$$$\int \frac{\sec {}^3\theta \mathrm{d}\theta }{2\sqrt{\tan \theta }}$$$$......$$$$$$

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