If integrating factor of $x\left(1-x^2\right) d y+\left(2 x^2 y-y-a x^3\right) d x=0$ is $e^{\int P d x}$, then P is :

$\frac{2 x^2-1}{x\left(1-x^2\right)}$

$x\left(1-x^2\right) \frac{d y}{d x}+2 x^2 y-y-a x^3=0 $

$\Rightarrow \frac{d y}{d x}+\frac{\left(2 x^2-1\right)}{x\left(1-x^2\right)} y=\frac{a x^3}{x\left(1-x^2\right)}$

$\Rightarrow P=\frac{2 x^2-1}{x\left(1-x^2\right)}$

Hence (4) is the correct answer.