∫((1+x^6 )/(1+x^8 ))dx


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Question Number 93410 by M±th+et+s last updated on 12/May/20

$\int \frac{1 + x^{6}}{1 + x^{8}} d x$
Commented by prakash jain last updated on 13/May/20
$Were you able to solve this. My method of solving by partial fraction/roots of unity is giving very long answer.$
$Were you able to solve this . My$ $method of solving by partial$ $fraction / roots of unity is giving$ $very long answer .$
Commented by M±th+et+s last updated on 13/May/20

$i {d i d n}^{'} t f i n d a n y i d e a s t o s o l v e t h i s$ $i p o s t e d i t l o o k i n g f o r s o l u t i o n s$
Commented by prakash jain last updated on 13/May/20

$ok . then i will continue working$ $on iy .$
Commented by M±th+et+s last updated on 13/May/20

$t h a n k y o u s o m u c h s i r$
Commented by mathmax by abdo last updated on 14/May/20

$c o m p l e x m e t h o d I = \int \frac{1 + x^{6}}{1 + x^{8}} d x l e t s o l v e z^{8} + 1 = 0 \Rightarrow$ $z^{8} = e^{i (2 k + 1) π} \Rightarrow t h e r o o t s a r e z_{k} = e^{\frac{i (2 k + 1) π}{8}}, k \in [[0, 7]]$ $\Rightarrow \frac{1 + x^{6}}{1 + x^{8}} = \frac{1 + x^{6}}{\prod_{k = 0}^{7} (x - z_{k})} = \sum_{k = 0}^{7} \frac{a_{k}}{x - z_{k}} w i t h a_{k} = \frac{1 + z_{k}^{6}}{8 z_{k}^{7}}$ $= - \frac{1}{8} z_{k} (1 + z_{k}^{6}) = - \frac{1}{8} (z_{k} + \frac{z_{k}^{8}}{z_{k}}) = - \frac{1}{8} (z - \frac{1}{z_{k}})$ $= - \frac{1}{8} (z_{k} - {\bar{z}}_{k}) = - \frac{1}{8} (2 i s i n (\frac{2 k + 1) π}{8})) = - \frac{i}{4} s i n (\frac{k π}{4} + \frac{π}{8}) \Rightarrow$ $\frac{1 + x^{6}}{1 + x^{8}} = - \frac{i}{4} \sum_{k = 0}^{7} \frac{s i n (\frac{k π}{4} + \frac{π}{8})}{x - z_{k}} \Rightarrow$ $\int \frac{1 + x^{6}}{1 + x^{8}} d x = - \frac{i}{4} \sum_{k = 0}^{7} s i n (\frac{k π}{4} + \frac{π}{8}) l n (x - z_{k}) + C$
Commented by prakash jain last updated on 15/May/20

$Thanks$
Commented by mathmax by abdo last updated on 15/May/20

$y o u a r e w e l c o m e s i r .$
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