Need help to make numerator diff of denominator in integration, eqn [5].

Solve diff eqn \frac {\delta y}{\delta x} = \frac {x^2 - y^2}{y^2 + x^2} ..........[1]

Let y = vx , where \, v = f(x)

\frac {\delta y}{\delta x} = v + x \frac {\delta v}{\delta x} .....[2]

substituting for y in [1], \frac {\delta y}{\delta x} = \frac {x^2 - v^2 x^2}{v^2 x^2 + x^2}

\frac {\delta y}{\delta x} = \frac {x^2(1 - v^2) }{x^2(v^2 + 1)}

\frac {\delta y}{\delta x} = \frac {1 - v^2 }{v^2 + 1} ........[3]

equating 2 and 3, \frac {1 - v^2 }{v^2 + 1} = v + x \frac {\delta v}{\delta x}

\frac {1 - v^2 }{v^2 + 1} - v = x \frac {\delta v}{\delta x}

\frac {1 - v^2 - v^3 - v}{v^2 + 1} = x \frac {\delta v}{\delta x}

\frac {v^2 + 1}{1 - v^2 - v^3 - v} = \frac {1}{x} \frac {\delta x}{\delta v}

\int \left ( \frac {v^2 + 1}{1 - v^2 - v^3 - v} \right ) \delta y = \int \frac {1}{x} \delta x .........[4]

comment : \frac {\delta(1 - v^2 - v^3 - v)}{\delta v} = 2v - 3v^2 -1

\therefore \, ???\int \left ( \frac {2v - 3v^2 -1}{1 - v^2 - v^3 - v} \right ) \delta y = \int \frac {1}{x} \delta x........[5]

Help: need to replace numerator on LHS of [4] with numerator on LHS of [5]

\frac {1}{2v - 3v^2 -1} \cdot \int \left ( \frac {2v - 3v^2 -1}{1 - v^2 - v^3 - v} \right ) \delta y = \int \frac {1}{x} \delta x how do I factor in v^2 + 1 in numerator in extreme left ?