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Question Number 187855 by Michaelfaraday last updated on 23/Feb/23

$$solve$$$$\int \frac{x^4+x^2+1}{2(x^2+1)}dx$$

Answered by cortano12 last updated on 23/Feb/23

$$\mathrm{I}=\frac{1}{2}\int \frac{{({\mathrm{x}}^2+1)}^2-{\mathrm{x}}^2}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$\mathrm{I}=\frac{1}{6}{\mathrm{x}}^3+\frac{1}{2}\mathrm{x}-\frac{1}{2}\int \frac{{\mathrm{x}}^2}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$=\frac{1}{6}{\mathrm{x}}^3+\frac{1}{2}\mathrm{x}-\frac{1}{2}\int \frac{\tan {}^2\theta }{\sec {}^2\theta }\sec {}^2\theta \mathrm{d}\theta ;\mathrm{x}=\tan \theta$$$$=\frac{1}{6}{\mathrm{x}}^3+\frac{1}{2}\mathrm{x}-\frac{1}{2}\int (\sec {}^2\theta -1)\mathrm{d}\theta$$$$=\frac{1}{6}{\mathrm{x}}^3+\frac{1}{2}\mathrm{x}-\frac{1}{2}\mathrm{x}+\frac{1}{2}\arctan \mathrm{x}+\mathrm{C}$$$$=\frac{1}{6}{\mathrm{x}}^3+\frac{1}{2}\arctan \mathrm{x}+\mathrm{C}$$

Commented by Michaelfaraday last updated on 25/Feb/23

$$thankssir$$

Answered by CElcedricjunior last updated on 23/Feb/23

$$\int \frac{{\mathit{x}}^4+{\mathit{x}}^2+1}{2({\mathit{x}}^2+1)}\mathit{d}\mathit{x}=\frac{1}{2}\int [\frac{{\mathit{x}}^2({\mathit{x}}^2+1)}{({\mathit{x}}^2+1)}+\frac{1}{({\mathit{x}}^2+1)}]\mathit{d}\mathit{x}$$$$=\frac{1}{2}\int [{\mathit{x}}^2+\frac{1}{{\mathit{x}}^2+1}]\mathit{d}\mathit{x}◼Moivre$$$$=\frac{1}{2}[\frac{1}{3}{\mathit{x}}^3+\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(\mathit{x}\right)]+\mathit{k}★\mathcal{C}edricjunior$$

Commented by Michaelfaraday last updated on 25/Feb/23

$$thankssir$$

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