dx-x-10-x-2-

Question Number 102470 by M±th+et+s last updated on 09/Jul/20

\int \frac{d x}{x^{10} + x^{2}}

Answered by Ar Brandon last updated on 09/Jul/20

I = \int \frac{dx}{x^{10} + x^{2}} = \int \frac{dx}{x^{2} (x^{8} + 1)}

\frac{1}{x^{2} (x^{8} + 1)} = \frac{ax + b}{x^{2}} + \frac{{cx}^{7} + {dx}^{6} + {ex}^{5} + {fx}^{4} + {gx}^{3} + {hx}^{2} + ix + j}{x^{8} + 1}

x = 0 \Rightarrow b = 1, a + c = 0, d + b = 0 \Rightarrow d = - 1

e = 0, f = 0, g = 0, h = 0, i = 0, j = 0, a = 0 \Rightarrow c = 0

\frac{1}{x^{2} (x^{8} + 1)} = \frac{1}{x^{2}} - \frac{x^{6}}{x^{8} + 1}

\Rightarrow I = \int {\frac{1}{x^{2}} - \frac{x^{6}}{x^{8} + 1}} dx = - \frac{1}{x} - \int \frac{x^{6}}{x^{8} + 1} dx

J = \int \frac{x^{6}}{x^{8} + 1} dx, x^{8} + 1 = 0 \Rightarrow x^{8} = e^{(2 k + 1) π i} \Rightarrow x_{k} = e^{\frac{(2 k + 1) π i}{8}}

\Rightarrow x^{8} + 1 = \underset{k = 0}{\prod^{7}} (x - x_{k})

\Rightarrow J = \int \frac{x^{6}}{\underset{k = 0}{\prod^{7}} (x - x_{k})} dx = \int \underset{k = 0}{\sum^{7}} \frac{a_{k}}{x - x_{k}} dx

a_{k} = \frac{x_{k}^{6}}{{8 x}_{k}^{7}} = \frac{x_{k}^{7}}{- 8} \Rightarrow J = - \frac{1}{8} \int \underset{k = 0}{\sum^{7}} \frac{x_{k}^{7} dx}{(x - x_{k})}

\Rightarrow J = - \frac{1}{8} \underset{k = 0}{\sum^{7}} x_{k}^{7} \ln (x - x_{k})

\Rightarrow \int \frac{dx}{x^{10} + x^{2}} = - \frac{1}{x} + \frac{1}{8} \underset{k = 0}{\sum^{7}} x_{k}^{7} \ln (x - x_{k}) + C

M r {M a t h m a x}^{'} s f a v o r i t e m e t h o d . H a h a!

I h o p e I^{'} v e d o n e i t i n t h e r i g h t w a y .

Commented by M±th+et+s last updated on 09/Jul/20

w e l l d o n e s i r

Commented by Ar Brandon last updated on 09/Jul/20

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Commented by mathmax by abdo last updated on 09/Jul/20

yes your answer is correct

Commented by floor(10²Eta[1]) last updated on 09/Jul/20

can you explain me how did you go from

here : \int \frac{x^{6}}{\underset{k = 0}{\prod^{7}} (x - x_{k})} dx to := \int \underset{k = 0}{\sum^{7}} \frac{a_{k}}{x - x_{k}} dx

and why a_{k} = \frac{x_{k}^{6}}{{8 x}_{k}^{7}}

Commented by Ar Brandon last updated on 09/Jul/20

Hi! Mathmax explained this to me in Q 86484 .

I guess you have a look at it .

Commented by Ar Brandon last updated on 09/Jul/20

i t s a l i k e t h e o r e m i f F = \frac{p (x)}{Q (x)} w i t h d e g p < d e g Q a n d

Q w i t h o u t r e a l r o o t s (\Rightarrow Q (x) = λ \prod_{i} (x - z_{i})) s o

F (x) = \sum_{i} \frac{a_{i}}{x - z_{i}} a n d a_{i} = \frac{p (z_{i})}{Q^{^{'}} (z_{i})}

Commented by floor(10²Eta[1]) last updated on 09/Jul/20

thanks! {that}^{'} s really nice, do you know where

i can see the demonstration of this ?

Commented by Ar Brandon last updated on 09/Jul/20

Thanks for your approval Sir Mathmax. ��

Commented by Ar Brandon last updated on 09/Jul/20

No idea, I don't know where you can find this demo. Maybe a little search on Google might help.��

Commented by mathmax by abdo last updated on 10/Jul/20

you are welcome sir