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 equal to

$\frac{2 x^2-a x^3}{x\left(1-x^2\right)}$ $2 x^3-1$ $\frac{2 x^2-1}{a x^3}$ $\frac{2 x^2-1}{x\left(1-x^2\right)}$

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

We have, $\frac{d y}{d x}+\frac{\left(2 x^2-1\right)}{x\left(1-x^2\right)} y=\frac{a x^2}{1-x^2}$ This is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$, where $P=\frac{2 x^2-1}{x\left(1-x^2\right)}$