# show me. thanks

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!!!!!!

### 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