Question Number 209542 by Spillover last updated on 13/Jul/24
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{{dx}}{{x}^{\mathrm{15}} −{x}^{\mathrm{11}} } \\ $$$$ \\ $$$$ \\ $$
Answered by Frix last updated on 14/Jul/24
$$\int\frac{{dx}}{{x}^{\mathrm{11}} \left({x}^{\mathrm{4}} −\mathrm{1}\right)}= \\ $$$$=\int\left(\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)}−\frac{{x}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{7}} }−\frac{\mathrm{1}}{{x}^{\mathrm{11}} }\right){dx}= \\ $$$$… \\ $$$$=\frac{\mathrm{15}{x}^{\mathrm{8}} +\mathrm{5}{x}^{\mathrm{4}} +\mathrm{3}}{\mathrm{30}{x}^{\mathrm{10}} }+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\frac{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{{x}^{\mathrm{2}} +\mathrm{1}}\:+{C} \\ $$
Commented by Spillover last updated on 14/Jul/24
$${great} \\ $$
Answered by Spillover last updated on 14/Jul/24
![∫(dx/(x^(15) −x^(11) )) t=x^2 xdx=(1/2)dt ∫(dx/(x^(15) −x^(11) ))=(1/2)∫(dx/(t^8 −t^6 ))=(1/2)∫(dx/(t^6 (t^2 −1))) u=(1/t) dt=−(1/u^2 )du (1/2)∫(dx/(t^6 (t^2 −1)))=(1/2)∫((−u^6 )/(1−u^2 ))du=(1/2)∫((1−u^6 −1)/(1−u^2 ))du (1/2)∫((1−u^6 −1)/(1−u^2 ))du=(1/2)∫((1−u^6 )/(1−u^2 ))du−(1/2)∫(du/(1−u^2 )) (1/2)∫((1−u^6 )/(1−u^2 ))du−(1/2)∫(du/(1−u^2 )) (1/2)∫(((1−u^2 )(1+u^2 +u^4 ))/(1−u^2 ))du−(1/2)∫(du/(1−u^2 )) (1/2)∫(1+u^2 +u^4 )du−(1/2)∫(du/(1−u^2 )) (1/2)[u+(u^3 /3)+(u^5 /5)]+(1/4)ln ((u−1)/(u+1)) but u=(1/x^2 ) (1/(2x^(2 ) ))+(1/(6x^6 ))+(1/(10x^(10) ))+(1/4)ln ((∣1−x^2 ∣)/(∣1+x^2 ∣))+A](https://www.tinkutara.com/question/Q209551.png)
$$\:\int\frac{{dx}}{{x}^{\mathrm{15}} −{x}^{\mathrm{11}} } \\ $$$${t}={x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{xdx}=\frac{\mathrm{1}}{\mathrm{2}}{dt} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{15}} −{x}^{\mathrm{11}} }=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{t}^{\mathrm{8}} −{t}^{\mathrm{6}} }=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{t}^{\mathrm{6}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$${u}=\frac{\mathrm{1}}{{t}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{dt}=−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }{du} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{t}^{\mathrm{6}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−{u}^{\mathrm{6}} }{\mathrm{1}−{u}^{\mathrm{2}} }{du}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−{u}^{\mathrm{6}} −\mathrm{1}}{\mathrm{1}−{u}^{\mathrm{2}} }{du} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−{u}^{\mathrm{6}} −\mathrm{1}}{\mathrm{1}−{u}^{\mathrm{2}} }{du}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−{u}^{\mathrm{6}} }{\mathrm{1}−{u}^{\mathrm{2}} }{du}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−{u}^{\mathrm{6}} }{\mathrm{1}−{u}^{\mathrm{2}} }{du}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{2}} +{u}^{\mathrm{4}} \right)}{\mathrm{1}−{u}^{\mathrm{2}} }{du}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+{u}^{\mathrm{2}} +{u}^{\mathrm{4}} \right){du}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[{u}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}\right]+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}} \\ $$$${but}\:\:\:\:\:\:\:\:\:\:\:{u}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}\:} }+\frac{\mathrm{1}}{\mathrm{6}{x}^{\mathrm{6}} }+\frac{\mathrm{1}}{\mathrm{10}{x}^{\mathrm{10}} }+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\frac{\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}{\mid\mathrm{1}+{x}^{\mathrm{2}} \mid}+{A} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$