\begin{array}{l}\text{The integrating factor of the differential equation}\mathrm{ x}\frac{dy}{dx}-2y=x^3\text{is}\end{array}

The correct option is 1/x².

To find the integrating factor of the given differential equation, we first need to write it in the standard form of a first-order linear differential equation.

The given differential equation is:

$$x\frac{dy}{dx}-2y=x^3$$

The standard form of a first-order linear differential equation is:

$$\frac{dy}{dx}+P(x)y=Q(x)$$

To convert the given equation into standard form, we divide every term by $x$ (where $x\ne 0$):

$$\frac{dy}{dx}-\frac{2}{x}y=x^2$$

By comparing this with the standard form, we can identify the coefficient function $P(x)$ as:

$$P(x)=-\frac{2}{x}$$

The integrating factor (I.F.) is calculated using the formula:

$$\text{I.F.}=e^{\int P(x)dx}$$

Substitute the value of $P(x)$ into the formula:

$$\text{I.F.}=e^{\int -\frac{2}{x}dx}$$

Evaluate the integral in the exponent:

$$\int -\frac{2}{x}dx=-2ln(x)$$

Using the logarithmic property $aln(b)=ln(b^a)$, we can rewrite the exponent as:

$$-2ln(x)=ln(x^{-2})$$

Now, substitute this back into the expression for the integrating factor:

$$\text{I.F.}=e^{ln(x^{-2})}$$

Since the exponential function and natural logarithm are inverse operations, $e^{ln(f(x))}=f(x)$. Therefore, we have:

$$\text{I.F.}=x^{-2}=\frac{1}{x^2}$$

Thus, the integrating factor is 1/x².