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Question Number 130778 by EDWIN88 last updated on 28/Jan/21

$$\int \frac{3x^2-1}{x^4-1}dx$$

Answered by bemath last updated on 29/Jan/21

$$\frac{{3\mathrm{x}}^2-1}{({\mathrm{x}}^2+1)({\mathrm{x}}^2-1)}=\frac{3({\mathrm{x}}^2-1)+2}{({\mathrm{x}}^2+1)({\mathrm{x}}^2-1)}$$$$\mathrm{I}=\int \frac{3}{{\mathrm{x}}^2+1}\mathrm{dx}+\int (\frac{1}{{\mathrm{x}}^2-1}-\frac{1}{{\mathrm{x}}^2+1})\mathrm{dx}$$$$\mathrm{I}={3\tan }^{-1}\left(\mathrm{x}\right)+\frac{1}{2}\int (\frac{1}{\mathrm{x}-1}-\frac{1}{\mathrm{x}+1})\mathrm{dx}-{\tan }^{-1}\left(\mathrm{x}\right)$$$$\mathrm{I}={2\tan }^{-1}\left(\mathrm{x}\right)+\frac{1}{2}\ln \mid \frac{\mathrm{x}-1}{\mathrm{x}+1}\mid +\mathrm{c}$$

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