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I-x-x-4-1-2-dx-

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Question Number 132070 by bramlexs22 last updated on 10/Feb/21
I=∫ (x/((x^4 −1)^2 )) dx ?
$$\mathrm{I}=\int\:\frac{\mathrm{x}}{\left(\mathrm{x}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx}\:? \\ $$
Answered by liberty last updated on 10/Feb/21
I= (1/2)∫ ((d(x^2 ))/(((x^2 )^2 −1)^2 )) = (1/2)∫ (dt/((t^2 −1)^2 )) I=(1/4)∫ ((t^2 +1)/((t^2 −1)^2 ))dt−(1/4)∫ ((t^2 −1)/((t^2 −1)^2 ))dt I= (1/4)∫ ((1+(1/t^2 ))/((t−(1/t))^2 ))dt−(1/4)∫(dt/(t^2 −1)) I=−(1/4). (1/(t−(1/t)))−(1/8)∫((1/(t−1))−(1/(t+1)))dt I=−(t/(4(t^2 −1)))−(1/8)ln ∣((t−1)/(t+1))∣+c I=−(x^2 /(4(x^4 −1)))−(1/8)ln ∣((x^2 −1)/(x^2 +1))∣+c
$$\mathrm{I}=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} \right)}{\left(\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}−\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\mathrm{I}=\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} }\mathrm{dt}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{I}=−\frac{\mathrm{1}}{\mathrm{4}}.\:\frac{\mathrm{1}}{\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}}−\frac{\mathrm{1}}{\mathrm{8}}\int\left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right)\mathrm{dt} \\ $$$$\mathrm{I}=−\frac{\mathrm{t}}{\mathrm{4}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid\frac{\mathrm{t}−\mathrm{1}}{\mathrm{t}+\mathrm{1}}\mid+\mathrm{c} \\ $$$$\mathrm{I}=−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{x}^{\mathrm{4}} −\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mid+\mathrm{c} \\ $$$$ \\ $$

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