The integrating factor of $\frac{{xdy}}{{dx}} - y = {x^4} - 3x$ is

Given that, $x\frac{{dy}}{{dx}} - y = {x^4} - 3x$

$\Rightarrow$ $\frac{{dy}}{{dx}} - \frac{y}{x} = {x^3} - 3$

Here, $P = - \frac{1}{x},Q = {x^3} - 3$

$\therefore \mid$ $IF = {e^{\int P dx}} = {e^{ - \int {\frac{1}{x}} dx}} = {e^{ - \log x}}$

$= \frac{1}{x}$