What is the solution for (u-2)/(u^2-2u+1)du=-1/x dx ?

The solution for (u-2)/(u^2-2u+1)du=-1/x dx is u(x)=1-1/(\W(-\frac{x){c_{1}})}

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(u-2)/(u^2-2u+1)du=-1/x dx

\frac{u - 2}{u ^{2} - 2 u + 1} d u = - \frac{1}{x} d x

u (x) = 1 - \frac{1}{W ( - \frac{x}{c _{1}} )}

Solution steps

\frac{u ( x ) - 2}{u ( x ) ^{2} - 2 u ( x ) + 1} d u (x) = - \frac{1}{x} d x

Solve separable ODE:

u (x) = 1 - \frac{1}{W ( - \frac{x}{c _{1}} )}

u (x) = 1 - \frac{1}{W ( - \frac{x}{c _{1}} )}

f (x) = (9 x - 1)^{- \frac{1}{3}}, at x = 1

t, \frac{d y}{d t} = y^{3} - y^{2} - 12 y

f (x) = \frac{4 e ^{x} + 5 e ^{- x}}{6}

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 4})

\frac{d y}{d x} = \frac{x ^{3}}{y + 1}

What is the solution for (u-2)/(u^2-2u+1)du=-1/x dx ?
The solution for (u-2)/(u^2-2u+1)du=-1/x dx is u(x)=1-1/(\W(-\frac{x){c_{1}})}