show me. thanks

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# show me. thanks

Submitted by nona_nonomo on Fri, 10/29/2010 - 15:56

Forums:

is there a website which major in solving elementory Math (i.e. solve online high school mathematic problems). give me address, please. Thanks a lots!!!!!!

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## Versions

(v1) by nona_nonomo 2010-10-29

## Re: show me. thanks

Hi Nona,

The Eq.

x^12 + 1 = 0 (1)

has not real roots. So you have 6 complex plus its 6 conjugated roots. Use undetermined coefficients method to find out the constants in

x^6/(x^{12} + 1) = A_1/(x - w_1) + A_2/(x - w_2) + ... + A_{11}/(x - w_{11}) + A_{12}/(x - w_{12})

(Recall to separate real and imaginary parts).

Put

A_k = a_k + i.b_k,

where

k = 1, 2, ... , 12.

From (1),

x = (-1)^{1/12} = e^{i.(\pi + 2k.\pi)/12} = cos[(\pi/ + 2k.\pi)/12] + i.sin[(\pi + 2k.\pi)/12]

you get the w_k.

So,

\int[x^6/(x^{12} +1)]dx

= A_1.\log(x - w_1) + A_2.\log(x - w_2) + ... + A_{11}.\log(x - w_{11}) + A_{12}.\log(x - w_{12}).

By calling

w_k = u_k + i.v_k,

you have

x - w_k = (x - u_k) - i.v_k = r_k.e^{i.\theta_k},

where

r_k = \sqrt{(x - u_k)^2 + v_k^2},

and

tan(\theta_k) = -v_k/(x - u_k).

Therefore,

\log(x - w_k) = \log(r_k.e^{i.\theta_k}) = \log(r_k) + i.\theta_k = (1/2).\log[(x - u_k)^2 + v_k^2] - i.tan^{-1}[v_k/(x - u_k)],

for all k = 1, 2, ... , 12.

You must do a lot of calculations as these ones increase abruptly with the high exponents of your original rational function.

Please solve this integral and never ever solve another similar. That would be Sado-Masochism.

perucho