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Question Number 132070 by bramlexs22 last updated on 10/Feb/21

$$\mathrm{I}=\int \frac{\mathrm{x}}{{({\mathrm{x}}^4-1)}^2}\mathrm{dx}?$$

Answered by liberty last updated on 10/Feb/21

$$\mathrm{I}=\frac{1}{2}\int \frac{\mathrm{d}\left({\mathrm{x}}^2\right)}{{({\left({\mathrm{x}}^2\right)}^2-1)}^2}=\frac{1}{2}\int \frac{\mathrm{dt}}{{({\mathrm{t}}^2-1)}^2}$$$$\mathrm{I}=\frac{1}{4}\int \frac{{\mathrm{t}}^2+1}{{({\mathrm{t}}^2-1)}^2}\mathrm{dt}-\frac{1}{4}\int \frac{{\mathrm{t}}^2-1}{{({\mathrm{t}}^2-1)}^2}\mathrm{dt}$$$$\mathrm{I}=\frac{1}{4}\int \frac{1+\frac{1}{{\mathrm{t}}^2}}{{(\mathrm{t}-\frac{1}{\mathrm{t}})}^2}\mathrm{dt}-\frac{1}{4}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-1}$$$$\mathrm{I}=-\frac{1}{4}.\frac{1}{\mathrm{t}-\frac{1}{\mathrm{t}}}-\frac{1}{8}\int (\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1})\mathrm{dt}$$$$\mathrm{I}=-\frac{\mathrm{t}}{4({\mathrm{t}}^2-1)}-\frac{1}{8}\ln \mid \frac{\mathrm{t}-1}{\mathrm{t}+1}\mid +\mathrm{c}$$$$\mathrm{I}=-\frac{{\mathrm{x}}^2}{4({\mathrm{x}}^4-1)}-\frac{1}{8}\ln \mid \frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+1}\mid +\mathrm{c}$$$$$$

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