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Question Number 202388 by Calculusboy last updated on 25/Dec/23

$$\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t}:\int \frac{\mathit{d}\mathit{x}}{{\mathit{b}}^4+2{\mathit{a}\mathit{x}}^2+\mathit{c}}=\frac{{\mathit{t}\mathit{a}\mathit{n}}^{-1}\left(\frac{\sqrt{2}\sqrt{\mathit{a}}\mathit{x}}{\sqrt{\mathit{c}+{\mathit{b}}^4}}\right)}{\sqrt{2}\sqrt{\mathit{a}}\sqrt{\mathit{c}+{\mathit{b}}^4}}+\mathit{C}$$$$\mathit{i}\mathit{f}\mathit{a}\cdot (\mathit{c}+{\mathit{b}}^4)>0$$$$$$

Answered by witcher3 last updated on 26/Dec/23

$$\int \frac{\mathrm{dx}}{{\mathrm{b}}^4+\mathrm{c}+{2\mathrm{ax}}^2}=\int \frac{\mathrm{dx}}{2\mathrm{a}({\mathrm{x}}^2+\frac{{\mathrm{b}}^4+\mathrm{c}}{2\mathrm{a}})};{\beta }^2=\frac{{\mathrm{b}}^4+\mathrm{c}}{2\mathrm{a}}>0$$$$=\frac{1}{2\mathrm{a}}\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+{\beta }^2}=\frac{1}{2\mathrm{a}\beta }{\tan }^{-1}\left(\frac{\mathrm{x}}{\beta }\right)+\mathrm{k};\mathrm{k}\in \mathbb{R}$$$$=\frac{1}{2\mathrm{a}\sqrt{\frac{{\mathrm{b}}^4+\mathrm{c}}{2\mathrm{a}}}}{\tan }^{-1}\left(\mathrm{x}\sqrt{\frac{2\mathrm{a}}{{\mathrm{b}}^4+\mathrm{c}}}\right)+\mathrm{k}$$$$=\frac{1}{\sqrt{2}.\sqrt{\mathrm{a}}.\sqrt{{\mathrm{b}}^4+\mathrm{c}}}.{\tan }^{-1}\left(\frac{\mathrm{x}\sqrt{2}.\sqrt{\mathrm{a}}}{\sqrt{\mathrm{c}+{\mathrm{b}}^4}}\right)+\mathrm{k}$$

Commented by Calculusboy last updated on 26/Dec/23

$$\mathit{n}\mathit{i}\mathit{c}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}$$

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