$\displaystyle
\int \frac{dx}{x(1+2\sqrt{x}+\sqrt[3]{x})}=\int \frac{dx}{x(\sqrt{x}+1+\sqrt{x}+\sqrt[3]{x})}=
\int \frac{dx}{x(\sqrt{x}+\frac{(1-x)}{1-\sqrt[3]{x}})}
$
i got read of one root but instead i got another one
??
You should get $\displaystyle
\int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1}
$
You have to split into 2 parts
$\displaystyle \int\frac{-\frac{3}{2}u+\frac{1}{4}}{2u^2-u+1} = -\frac{3}{8}\int\frac{4u-1}{2u^2-u+1} - \frac{1}{8}\int\frac{1}{2u^2-u+1}$
The first part is integrable in log
The second part is integrable in Arctan after setting t=u-1/4 then $\displaystyle v = \frac{4t}{\sqrt{7}}$
$\displaystyle
\frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}\\
$
$\displaystyle
A(u+1)(2u^2-u+1)-Bu(2u^2-u+1)+(Cu+D)(u^2+u)=1\\
$
$\displaystyle
u^3:\\
$
$\displaystyle
2A-2B+C=0\\
$
$\displaystyle
u^2:\\
$
$\displaystyle
A+B+C+D=0\\
u:\\
$
$\displaystyle
-B+D=0\\
$
1:
$\displaystyle
A=1
$
The denominator factorises to $\displaystyle u(1+2u^3+u^2) = u(u+1)(2u^2 - u +1)$.
Thus, The fraction becomes $\displaystyle \frac{1}{u(1+2u^3+u^2)} = \frac{A}{u} + \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}$.
You have a sign error. $\displaystyle B$ is $\displaystyle +$ve, and not $\displaystyle -$ve. Your formula produced are correct, just a sign error has made them slightly differ.
You can continue to use gaussian elimination after you've made the correction.
Your equations are correct.
There is a much easier way :
$\displaystyle \frac{1}{u(1+2u^3+u^2)} = \frac{1}{u(u+1)(2u^2-u+1)} = \frac{A}{u} - \frac{B}{u+1} + \frac{Cu+D}{2u^2-u+1}$
Multiply both sides by u
$\displaystyle \frac{1}{(u+1)(2u^2-u+1)} = A - \frac{Bu}{u+1} + \frac{u(Cu+D)}{2u^2-u+1}$
And set u to 0
$\displaystyle 1 = A$
Multiply both sides by u+1
$\displaystyle \frac{1}{u(2u^2-u+1)} = \frac{A(u+1)}{u} - B + \frac{(u+1)(Cu+D)}{2u^2-u+1}$
And set u to -1
$\displaystyle -\frac{1}{4} = - B$