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Question Number 211578 by MrGaster last updated on 13/Sep/24

$$$$$${\int }_0^{+\mathrm{\infty }}\frac{\mathit{x}}{\sqrt{1+{\mathit{x}}^4}}\mathit{d}\mathit{x}.$$$$$$

Answered by lepuissantcedricjunior last updated on 13/Sep/24

$${\int }_0^{\mathrm{\infty }}\frac{\mathit{x}\mathit{d}\mathit{x}}{\sqrt{1+{\mathit{x}}^4}}=\mathit{k}$$$$\mathit{k}={\int }_0^{\mathrm{\infty }}\frac{\mathit{d}\left({\mathit{x}}^2\right)}{2\sqrt{1+{\left({\mathit{x}}^2\right)}^2}}\stackrel{{\mathit{x}}^2=\mathit{t}\mathit{a}\mathit{n}\mathit{t}=>\mathit{d}\left({\mathit{x}}^2\right)=(1+{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{t})\mathit{d}\mathit{t}}{}$$$$\mathit{k}={\int }_0^{\mathit{\pi }/2}\frac{1+{\mathit{t}\mathit{a}\mathit{n}}^2\mathit{t}}{\frac{1}{\mathit{c}\mathit{o}\mathit{s}\mathit{t}}}={\int }_0^{\frac{\mathit{\pi }}{2}}(\mathit{c}\mathit{o}\mathit{s}\mathit{t}+\frac{{\mathit{s}\mathit{i}\mathit{n}}^2}{\mathit{c}\mathit{o}\mathit{s}\mathit{t}})\mathit{d}\mathit{t}$$$$\mathit{k}=1+{\int }_0^{\frac{\mathit{\pi }}{2}}(\frac{1}{\mathit{c}\mathit{o}\mathit{s}\mathit{t}}-\mathit{c}\mathit{o}\mathit{s}\mathit{t})\mathit{d}\mathit{t}$$$$\mathit{k}={\int }_0^{\frac{\mathit{\pi }}{2}}\left(\frac{1}{2}(\frac{\mathit{c}\mathit{o}\mathit{s}\mathit{t}}{1-\mathit{s}\mathit{i}\mathit{n}\mathit{t}}+\frac{\mathit{c}\mathit{o}\mathit{s}\mathit{t}}{1+\mathit{s}\mathit{i}\mathit{n}\mathit{t}})\right)\mathit{d}\mathit{t}$$$$\mathit{k}=\frac{1}{2}{\left[\mathit{l}\mathit{n}\left(\frac{1+\mathit{s}\mathit{i}\mathit{n}\mathit{t}}{1-\mathit{s}\mathit{i}\mathit{n}\mathit{t}}\right)\right]}_0^{\frac{\mathit{\pi }}{2}}=-\mathrm{\infty }$$

Answered by Frix last updated on 13/Sep/24

$$\int \frac{x}{\sqrt{1+x^4}}dx\stackrel{[t=x^2+\sqrt{x^4+1}]}{=}\frac{1}{2}\int \frac{dt}{t}=\frac{\ln t}{2}=$$$$=\frac{\ln (x^2+\sqrt{x^4+1})}{2}+C$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{x}{\sqrt{1+x^4}}dx\mathrm{diverges}$$

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