What is the solution for u^'-2/x \H(x)=-2/x ?

The solution for u^'-2/x \H(x)=-2/x is u(x)=integral of (-2+2\H(x))/x

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u^'-2/x \H(x)=-2/x

u^{^{'}} - \frac{2}{x} H (x) = - \frac{2}{x}

u (x) = \int \frac{- 2 + 2 H ( x )}{x} d x

Solution steps

u^{'} (x) - \frac{2}{x} H (x) = - \frac{2}{x}

Isolate

u^{'} (x) : u^{'} (x) = \frac{- 2 + 2 H ( x )}{x}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

u (x) = \int \frac{- 2 + 2 H ( x )}{x} d x

u (x) = \int \frac{- 2 + 2 H ( x )}{x} d x

What is the solution for u^'-2/x \H(x)=-2/x ?
The solution for u^'-2/x \H(x)=-2/x is u(x)=integral of (-2+2\H(x))/x