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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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