integral (x^2+1)/[(x^2+x+1)^2 + (x-2)^2]
what i tried is :
A/(x-2) + B/(x-2)^2 + (Cx+D)/(x^2+x+1) + (Ex+F)/(x^2+x+1)^2
then try to find ABCDEF
its seems too long and complicated is there any better way to solve it pls
thank u
integral (x^2+1)/[(x^2+x+1)^2 + (x-2)^2]
what i tried is :
A/(x-2) + B/(x-2)^2 + (Cx+D)/(x^2+x+1) + (Ex+F)/(x^2+x+1)^2
then try to find ABCDEF
its seems too long and complicated is there any better way to solve it pls
thank u
$\displaystyle \int \frac{x^2+1}{(x^2+x+1)^2(x-2)^2}\,dx$
$\displaystyle \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra c{E}{x-2}+\frac{F}{(x-2)^2}$.
Solve the identity for $\displaystyle A, B, C, D, E, F$.
This is going to be one ugly integral.
$\displaystyle \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra c{E}{x-2}+\frac{F}{(x-2)^2}$
Multiply both sides by $\displaystyle (x-2)^2$
$\displaystyle \frac{x^2+1}{(x^2+x+1)^2} = \frac{(Ax+B)(x-2)^2}{x^2+x+1}+\frac{(Cx+D)(x-2)^2}{(x^2+x+1)^2}+E(x-2)+F$
And set x to 2
$\displaystyle \frac{2^2+1}{(2^2+2+1)^2} = F = \frac{5}{49}$
There are no simple ways to obtain the remaining terms in this problem.
All techniques are more or less the same.
You can just multiply out and set the two polynominals equal to each other.
Then if you wish you can use matrices to solve for A...F.
Or one idea I never see people do, but it works.
Just let x=0,1,2,3,4, what ever it takes.
This too creates 6 equations and 6 unknows.
BUT you have no choice in this.
You can only steal the F.
You can also use complex numbers
$\displaystyle \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra c{E}{x-2}+\frac{F}{(x-2)^2}$
Multiply both sides by $\displaystyle (x^2+x+1)^2$
$\displaystyle \frac{x^2+1}{(x-2)^2} = (Ax+B)(x^2+x+1)+Cx+D+\frac{E(x^2+x+1)^2}{x-2}+\frac{F(x^2+x+1)^2}{(x-2)^2}$
And set x to j where $\displaystyle j=e^{\frac{2i\pi}{3}}$
$\displaystyle \frac{j^2+1}{(j-2)^2} = Cj+D$
Using $\displaystyle j^2=-j-1$ you can find C and D easily
For the other unknowns you can use
$\displaystyle \frac{x^2+1}{(x^2+x+1)^2(x-2)^2} = \frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}+\fra c{E}{x-2}+\frac{F}{(x-2)^2}$
with x=0 then x=1 then x=-1
which give the 3 last equations
This is a little bit work but it can be done !
thanks a lot but i cant do it this way it seems more complex i tried this but dont know if its correct or no .
i expanded the whole equation and then organized it to this form
A(x^5-3x^4+x^3+4x) + B(x^4-3x^3+x^2+4) + C(x^3-4x^2+4x) +D(x^2-4x+4) +E(x^5-x^3-4x^2+5x-2) +F(x^4+2x^3+3x^2+2x+1) = x^2 + 1
from this i got the following
A+E =0 (for x^5)
-3A+B+F =0 (for x^4)
A-3B+C-E+2F =0 (for x^3)
B-4C+D-4E+3F =1 (for x^2)
4A+4C-4D+5E+2F=0 (for x)
4B+4D-2E+F =1 (for cnst)
ok i already got E as u explained (E=5/7)
and F=(5/49)
and started to substitute
in the equations above so i got
A=-E= -5/7
B= -35/49
is that right or just a mess
thank u