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Question Number 108325 by john santu last updated on 16/Aug/20

$$\frac{⊸JS⊸}{----}$$$$\int \frac{dx}{x^8(x^2+1)}=?$$

Commented by john santu last updated on 16/Aug/20

$$\frac{⊸JS⊸}{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}$$$$\int \frac{dx}{x^8(x^2+1)}=\int (\frac{1}{x^2+1}-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\frac{1}{x^8})dx$$$$={\tan }^{-1}\left(x\right)+\frac{1}{x}-\frac{1}{3x^3}+\frac{1}{5x^5}-\frac{1}{7x^7}+c$$

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