Question Number 93410 by M±th+et+s last updated on 12/May/20
$$\int\frac{\mathrm{1}+{x}^{\mathrm{6}} }{\mathrm{1}+{x}^{\mathrm{8}} }{dx} \\ $$
Commented by prakash jain last updated on 13/May/20
$$\mathrm{Were}\:\mathrm{you}\:\mathrm{able}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}.\:\mathrm{My}\: \\ $$$$\mathrm{method}\:\mathrm{of}\:\mathrm{solving}\:\mathrm{by}\:\mathrm{partial} \\ $$$$\mathrm{fraction}/\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{is}\:\mathrm{giving} \\ $$$$\mathrm{very}\:\mathrm{long}\:\mathrm{answer}. \\ $$
Commented by M±th+et+s last updated on 13/May/20
$${i}\:{didn}'{t}\:{find}\:{any}\:{ideas}\:{to}\:{solve}\:{this} \\ $$$${i}\:{posted}\:{it}\:{looking}\:{for}\:{solutions} \\ $$
Commented by prakash jain last updated on 13/May/20
$$\mathrm{ok}.\:\mathrm{then}\:\mathrm{i}\:\mathrm{will}\:\mathrm{continue}\:\mathrm{working} \\ $$$$\mathrm{on}\:\mathrm{iy}. \\ $$
Commented by M±th+et+s last updated on 13/May/20
$${thank}\:{you}\:{so}\:{much}\:{sir}\: \\ $$
Commented by mathmax by abdo last updated on 14/May/20
![complex method I =∫ ((1+x^6 )/(1+x^8 ))dx let solve z^8 +1 =0 ⇒ z^8 =e^(i(2k+1)π) ⇒ the roots are z_k =e^((i(2k+1)π)/8) ,k ∈[[0,7]] ⇒((1+x^6 )/(1+x^8 )) =((1+x^6 )/(Π_(k=0) ^7 (x−z_k ))) =Σ_(k=0) ^7 (a_k /(x−z_k )) with a_k =((1+z_k ^6 )/(8 z_k ^7 )) =−(1/8)z_k (1+z_k ^6 ) =−(1/8)(z_k +(z_k ^8 /z_k )) =−(1/8)(z−(1/z_k )) =−(1/8)(z_k −z_k ^− ) =−(1/8)(2i sin(((2k+1)π)/8))) =−(i/4)sin(((kπ)/4) +(π/8)) ⇒ ((1+x^6 )/(1+x^8 )) =−(i/4)Σ_(k=0) ^7 ((sin(((kπ)/4)+(π/8)))/(x−z_k )) ⇒ ∫ ((1+x^6 )/(1+x^8 ))dx =−(i/4) Σ_(k=0) ^7 sin(((kπ)/4) +(π/8))ln(x−z_k ) +C](https://www.tinkutara.com/question/Q93616.png)
$${complex}\:{method}\:\:{I}\:=\int\:\:\:\frac{\mathrm{1}+{x}^{\mathrm{6}} }{\mathrm{1}+{x}^{\mathrm{8}} }{dx}\:\:{let}\:{solve}\:{z}^{\mathrm{8}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow \\ $$$${z}^{\mathrm{8}} \:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\Rightarrow\:{the}\:{roots}\:{are}\:{z}_{{k}} ={e}^{\frac{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{8}}} \:\:\:\:,{k}\:\in\left[\left[\mathrm{0},\mathrm{7}\right]\right] \\ $$$$\Rightarrow\frac{\mathrm{1}+{x}^{\mathrm{6}} }{\mathrm{1}+{x}^{\mathrm{8}} }\:=\frac{\mathrm{1}+{x}^{\mathrm{6}} }{\prod_{{k}=\mathrm{0}} ^{\mathrm{7}} \left({x}−{z}_{{k}} \right)}\:=\sum_{{k}=\mathrm{0}} ^{\mathrm{7}} \:\:\frac{{a}_{{k}} }{{x}−{z}_{{k}} }\:\:{with}\:{a}_{{k}} =\frac{\mathrm{1}+{z}_{{k}} ^{\mathrm{6}} }{\mathrm{8}\:{z}_{{k}} ^{\mathrm{7}} }\: \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}{z}_{{k}} \left(\mathrm{1}+{z}_{{k}} ^{\mathrm{6}} \right)\:=−\frac{\mathrm{1}}{\mathrm{8}}\left({z}_{{k}} \:+\frac{{z}_{{k}} ^{\mathrm{8}} }{{z}_{{k}} }\right)\:=−\frac{\mathrm{1}}{\mathrm{8}}\left({z}−\frac{\mathrm{1}}{{z}_{{k}} }\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\left({z}_{{k}} −\overset{−} {{z}}_{{k}} \right)\:=−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{i}\:{sin}\left(\frac{\left.\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{8}}\right)\right)\:=−\frac{{i}}{\mathrm{4}}{sin}\left(\frac{{k}\pi}{\mathrm{4}}\:+\frac{\pi}{\mathrm{8}}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{1}+{x}^{\mathrm{6}} }{\mathrm{1}+{x}^{\mathrm{8}} }\:=−\frac{{i}}{\mathrm{4}}\sum_{{k}=\mathrm{0}} ^{\mathrm{7}} \:\frac{{sin}\left(\frac{{k}\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}\right)}{{x}−{z}_{{k}} }\:\Rightarrow \\ $$$$\int\:\:\frac{\mathrm{1}+{x}^{\mathrm{6}} }{\mathrm{1}+{x}^{\mathrm{8}} }{dx}\:=−\frac{{i}}{\mathrm{4}}\:\sum_{{k}=\mathrm{0}} ^{\mathrm{7}} \:{sin}\left(\frac{{k}\pi}{\mathrm{4}}\:+\frac{\pi}{\mathrm{8}}\right){ln}\left({x}−{z}_{{k}} \right)\:+{C} \\ $$$$ \\ $$
Commented by prakash jain last updated on 15/May/20
$$\mathrm{Thanks} \\ $$
Commented by mathmax by abdo last updated on 15/May/20
$${you}\:{are}\:{welcome}\:{sir}. \\ $$