Question Number 86258 by lémùst last updated on 27/Mar/20
$${calculate}\:{I}=\int_{\mathrm{1}} ^{+\infty} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$
Commented by mathmax by abdo last updated on 27/Mar/20
![I =∫_1 ^(+∞) ((x^2 −1)/(x^4 −x^2 +1))dx (complex method) x^4 −x^2 +1 =0 →t^2 −t+1 =0 (t=x^2 ) Δ=1−4=−3 ⇒t_1 =((1+i(√3))/2)=e^((iπ)/6) and t_2 =((1−i(√3))/2)=e^(−((iπ)/6)) ⇒((x^2 −1)/(x^4 −x^2 +1)) =((x^2 −1)/((x^2 −e^((iπ)/6) )(x^2 −e^(−((iπ)/6)) ))) =((x^2 −1)/(2isin((π/6)))){(1/(x^2 −e^((iπ)/6) ))−(1/(x^2 −e^(−((iπ)/6)) ))} =−i(((x^2 −1)/(x^2 −e^((iπ)/6) )) −((x^2 −1)/(x^2 −e^(−((iπ)/6)) ))) =−i(((x^2 −e^((iπ)/6) +e^((iπ)/6) −1)/(x^2 −e^((iπ)/6) ))−((x^2 −e^(−((iπ)/6)) +e^(−((iπ)/6)) −1)/(x^2 −e^(−((iπ)/6)) ))) =−i{((e^((iπ)/6) −1)/(x^2 −e^((iπ)/6) ))−((e^(−((iπ)/6)) −1)/(x^2 −e^(−((iπ)/6)) ))} ⇒I=i(e^(−((iπ)/6)) −1)∫_1 ^(+∞) (dx/(x^2 −e^(−((iπ)/6)) )) −i(e^((iπ)/6) −1)∫_1 ^(+∞) (dx/(x^2 −e^((iπ)/6) )) ∫_1 ^(+∞) (dx/(x^2 −e^((iπ)/6) )) =∫_1 ^(+∞) (dx/((x−e^((iπ)/(12)) )(x+e^((iπ)/(12)) ))) =(1/2)e^(−((iπ)/(12))) ∫_1 ^(+∞) ((1/(x−e^((iπ)/(12)) ))−(1/(x+e^((iπ)/(12)) )))dx =(1/2)e^(−((iπ)/(12))) [ln(((x−e^((iπ)/(12)) )/(x+e^((iπ)/(12)) )))]_1 ^(+∞) =(1/2)e^(−((iπ)/(12))) {−ln(((1−e^((iπ)/(12)) )/(1+e^((iπ)/(12)) )))} ...becontinued...](https://www.tinkutara.com/question/Q86264.png)
$${I}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\:\:\left({complex}\:{method}\right) \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\rightarrow{t}^{\mathrm{2}} −{t}+\mathrm{1}\:=\mathrm{0}\:\:\left({t}={x}^{\mathrm{2}} \right) \\ $$$$\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}\:\Rightarrow{t}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{\frac{{i}\pi}{\mathrm{6}}} \:\:{and}\:{t}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{−\frac{{i}\pi}{\mathrm{6}}} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\left({x}^{\mathrm{2}} \:−{e}^{−\frac{{i}\pi}{\mathrm{6}}} \right)}\:=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{6}}\right)}\left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{6}}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:−{e}^{−\frac{{i}\pi}{\mathrm{6}}} }\right\} \\ $$$$=−{i}\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{6}}} }\:−\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:−{e}^{−\frac{{i}\pi}{\mathrm{6}}} }\right)\:=−{i}\left(\frac{{x}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{6}}} \:+{e}^{\frac{{i}\pi}{\mathrm{6}}} −\mathrm{1}}{{x}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{6}}} }−\frac{{x}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{6}}} +{e}^{−\frac{{i}\pi}{\mathrm{6}}} −\mathrm{1}}{{x}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{6}}} }\right) \\ $$$$=−{i}\left\{\frac{{e}^{\frac{{i}\pi}{\mathrm{6}}} −\mathrm{1}}{{x}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{6}}} }−\frac{{e}^{−\frac{{i}\pi}{\mathrm{6}}} −\mathrm{1}}{{x}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{6}}} }\right\}\:\Rightarrow{I}={i}\left({e}^{−\frac{{i}\pi}{\mathrm{6}}} −\mathrm{1}\right)\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −{e}^{−\frac{{i}\pi}{\mathrm{6}}} } \\ $$$$−{i}\left({e}^{\frac{{i}\pi}{\mathrm{6}}} −\mathrm{1}\right)\int_{\mathrm{1}} ^{+\infty} \frac{{dx}}{{x}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{6}}} } \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −{e}^{\frac{{i}\pi}{\mathrm{6}}} }\:=\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\left({x}−{e}^{\frac{{i}\pi}{\mathrm{12}}} \right)\left({x}+{e}^{\frac{{i}\pi}{\mathrm{12}}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{12}}} \int_{\mathrm{1}} ^{+\infty} \:\left(\frac{\mathrm{1}}{{x}−{e}^{\frac{{i}\pi}{\mathrm{12}}} }−\frac{\mathrm{1}}{{x}+{e}^{\frac{{i}\pi}{\mathrm{12}}} }\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{12}}} \left[{ln}\left(\frac{{x}−{e}^{\frac{{i}\pi}{\mathrm{12}}} }{{x}+{e}^{\frac{{i}\pi}{\mathrm{12}}} }\right)\right]_{\mathrm{1}} ^{+\infty} \:=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{12}}} \left\{−{ln}\left(\frac{\mathrm{1}−{e}^{\frac{{i}\pi}{\mathrm{12}}} }{\mathrm{1}+{e}^{\frac{{i}\pi}{\mathrm{12}}} }\right)\right\} \\ $$$$…{becontinued}… \\ $$
Commented by lémùst last updated on 28/Mar/20
$${sir}\:{how}\:{do}\:{you}\:{calculate}\:{the}\:{logarith}\:{of}\:{a} \\ $$$${complex}\:? \\ $$
Commented by mathmax by abdo last updated on 31/Mar/20
$${ln}\left({x}+{iy}\right)\:={ln}\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{e}^{{i}\:{arctan}\left(\frac{{y}}{{x}}\right)} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)+{iarctan}\left(\frac{{y}}{{x}}\right)\:\:\left({principle}\:{determination}\:{of}\:{ln}\right) \\ $$
Answered by TANMAY PANACEA. last updated on 28/Mar/20
$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\int\frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{3}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{3}}}\right) \\ $$$${useing}\:{formula}\:\int\frac{{dp}}{{p}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by lémùst last updated on 28/Mar/20
$${thanks}\:{sir}\: \\ $$
Commented by TANMAY PANACEA. last updated on 28/Mar/20
$${most}\:{welcome}… \\ $$