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Question Number 88206 by jagoll last updated on 09/Apr/20

$$\int \frac{\mathrm{x}+{\mathrm{x}}^3}{1+{\mathrm{x}}^4}\mathrm{dx}$$

Answered by john santu last updated on 09/Apr/20

$$=\int \frac{x}{1+x^4}dx+\int \frac{x^3}{1+x^4}dx$$$$=\frac{1}{2}\int \frac{2x}{1+x^4}dx+\frac{1}{4}\int \frac{4x^3}{1+x^4}dx$$$$=\frac{1}{2}\int \frac{d\left(x^2\right)}{1+{\left(x^2\right)}^2}+\frac{1}{4}\int \frac{d(1+x^4)}{1+x^4}$$$$=\frac{1}{4}\ln (1+x^4)+\frac{1}{2}\int \frac{du}{1+u^2},[u=x^2]$$$$=\frac{1}{4}\ln (1+x^4)+\frac{1}{2}arc\tan \left(x^2\right)+c$$

Commented by jagoll last updated on 09/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{mr}$$

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