Question Number 46221 by Meritguide1234 last updated on 22/Oct/18
Answered by MJS last updated on 22/Oct/18
$$\mathrm{I}\:\mathrm{solved}\:\mathrm{it}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{too}\:\mathrm{tired}\:\mathrm{to}\:\mathrm{type}\:\mathrm{it} \\ $$
Commented by Meritguide1234 last updated on 23/Oct/18
$${post}\:{your}\:{two}\:{or}\:{three}\:{line}\:{approach} \\ $$
Commented by MJS last updated on 23/Oct/18
$$\mathrm{will}\:\mathrm{type}\:\mathrm{completely}\:\mathrm{when}\:\mathrm{I}'\mathrm{m}\:\mathrm{wake}\:\mathrm{enough} \\ $$
Answered by MJS last updated on 23/Oct/18
![∫((√(x^2 +1))/(x(√(x^4 +1))(√(x^8 −1))))dx=∫(dx/(x(x^4 +1)(√(x^2 −1))))= [t=(√(x^2 −1)) → dx=((√(x^2 −1))/x)dt] =∫(dt/((t^2 +1)(t^4 +2t^2 +2)))=∫(dt/(t^2 +1))−∫((t^2 +1)/(t^4 +2t^2 +2))dt ∫(dt/(t^2 +1))=arctan t =arctan (√(x^2 −1)) ∫((t^2 +1)/(t^4 +2t^2 +2))dt=∫((t^2 +1)/((t^2 −at+b)(t^2 +at+b)))dt= [a=(√(−2+2(√2))); b=(√2)] =(1/(2ab))∫(((b−1)t+a)/(t^2 −at+b))dt−(1/(2ab))∫(((b−1)t−a)/(t^2 +at+b))dt= =((b−1)/(4ab))∫((2t−a)/(t^2 −at+b))dt+((b+1)/(4b))∫(dt/(t^2 −at+b))−((b−1)/(4ab))∫((2t+a)/(t^2 +at+b))dt+((b+1)/(4b))∫(dt/(t^2 +at+b))= =((b−1)/(4ab))ln (t^2 −at+b) +((b+1)/(2b(√(4b−a^2 ))))arctan ((2t−a)/( (√(4b−a^2 )))) −((b−1)/(4ab))ln (t^2 +at+b) +((b+1)/(2b(√(4b−a^2 ))))arctan ((2t+a)/( (√(4b−a^2 ))))= =((b−1)/(4ab))ln ∣((x^2 +b−1−a(√(x^2 −1)))/(x^2 +b−1+a(√(x^2 −1))))∣ +((b+1)/(2b(√(4b−a^2 ))))(arctan ((a+2(√(x^2 −1)))/( (√(4b−a^2 )))) −arctan ((a−2(√(x^2 −1)))/( (√(4b−a^2 ))))) ∫((√(x^2 +1))/(x(√(x^4 +1))(√(x^8 −1))))dx= =((b−1)/(4ab))ln ∣((x^2 +b−1+a(√(x^2 −1)))/(x^2 +b−1−a(√(x^2 −1))))∣ +((b+1)/(2b(√(4b−a^2 ))))(arctan ((a−2(√(x^2 −1)))/( (√(4b−a^2 )))) −arctan ((a+2(√(x^2 −1)))/( (√(4b−a^2 )))))+arctan (√(x^2 −1)) +C with a=(√(−2+2(√2))); b=(√2)](https://www.tinkutara.com/question/Q46258.png)
$$\int\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\sqrt{{x}^{\mathrm{8}} −\mathrm{1}}}{dx}=\int\frac{{dx}}{{x}\left({x}^{\mathrm{4}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}}{dt}\right] \\ $$$$=\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}\right)}=\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}−\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}{dt} \\ $$$$ \\ $$$$\:\:\:\:\:\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{arctan}\:{t}\:=\mathrm{arctan}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$ \\ $$$$\:\:\:\:\:\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}{dt}=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} −{at}+{b}\right)\left({t}^{\mathrm{2}} +{at}+{b}\right)}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{a}=\sqrt{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}};\:{b}=\sqrt{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}{ab}}\int\frac{\left({b}−\mathrm{1}\right){t}+{a}}{{t}^{\mathrm{2}} −{at}+{b}}{dt}−\frac{\mathrm{1}}{\mathrm{2}{ab}}\int\frac{\left({b}−\mathrm{1}\right){t}−{a}}{{t}^{\mathrm{2}} +{at}+{b}}{dt}= \\ $$$$\:\:\:\:\:=\frac{{b}−\mathrm{1}}{\mathrm{4}{ab}}\int\frac{\mathrm{2}{t}−{a}}{{t}^{\mathrm{2}} −{at}+{b}}{dt}+\frac{{b}+\mathrm{1}}{\mathrm{4}{b}}\int\frac{{dt}}{{t}^{\mathrm{2}} −{at}+{b}}−\frac{{b}−\mathrm{1}}{\mathrm{4}{ab}}\int\frac{\mathrm{2}{t}+{a}}{{t}^{\mathrm{2}} +{at}+{b}}{dt}+\frac{{b}+\mathrm{1}}{\mathrm{4}{b}}\int\frac{{dt}}{{t}^{\mathrm{2}} +{at}+{b}}= \\ $$$$\:\:\:\:\:=\frac{{b}−\mathrm{1}}{\mathrm{4}{ab}}\mathrm{ln}\:\left({t}^{\mathrm{2}} −{at}+{b}\right)\:+\frac{{b}+\mathrm{1}}{\mathrm{2}{b}\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\mathrm{arctan}\:\frac{\mathrm{2}{t}−{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\:−\frac{{b}−\mathrm{1}}{\mathrm{4}{ab}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +{at}+{b}\right)\:+\frac{{b}+\mathrm{1}}{\mathrm{2}{b}\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\mathrm{arctan}\:\frac{\mathrm{2}{t}+{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\:=\frac{{b}−\mathrm{1}}{\mathrm{4}{ab}}\mathrm{ln}\:\mid\frac{{x}^{\mathrm{2}} +{b}−\mathrm{1}−{a}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}^{\mathrm{2}} +{b}−\mathrm{1}+{a}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\mid\:+\frac{{b}+\mathrm{1}}{\mathrm{2}{b}\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\left(\mathrm{arctan}\:\frac{{a}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\:−\mathrm{arctan}\:\frac{{a}−\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\right) \\ $$$$ \\ $$$$\int\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\sqrt{{x}^{\mathrm{8}} −\mathrm{1}}}{dx}= \\ $$$$=\frac{{b}−\mathrm{1}}{\mathrm{4}{ab}}\mathrm{ln}\:\mid\frac{{x}^{\mathrm{2}} +{b}−\mathrm{1}+{a}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}^{\mathrm{2}} +{b}−\mathrm{1}−{a}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\mid\:+\frac{{b}+\mathrm{1}}{\mathrm{2}{b}\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\left(\mathrm{arctan}\:\frac{{a}−\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\:−\mathrm{arctan}\:\frac{{a}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\right)+\mathrm{arctan}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:+{C} \\ $$$$\mathrm{with}\:{a}=\sqrt{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}};\:{b}=\sqrt{\mathrm{2}} \\ $$
Commented by MJS last updated on 23/Oct/18
$$…\mathrm{please}\:\mathrm{check},\:\mathrm{I}\:\mathrm{already}\:\mathrm{corrected}\:\mathrm{some}\:\mathrm{typos} \\ $$
Commented by maxmathsup by imad last updated on 23/Oct/18
$${thank}\:{you}\:{sir}\:{you}\:{have}\:{played}\:{a}\:{criket}\:{match}\:{with}\:{this}\:{integral}… \\ $$
Commented by MJS last updated on 23/Oct/18
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$
Commented by Meritguide1234 last updated on 25/Oct/18
