An integrating factor of the differential equation
`(1+y+x^(2)y) dx + ( x +x^(3))dy = 0` is

Options

A. log x
B. x
C. `e^(x)`
D. `1/x`

Explanation

To find the integrating factor of the given differential equation

\[ (1 + y + x^2 y) \, dx + (x + x^3) \, dy = 0, \]

we will follow these steps:

### Step 1: Rewrite the Equation
We want to express the given differential equation in the standard linear form:

\[ \frac{dy}{dx} + P(x)y = Q(x). \]

Starting with the original equation, we can rearrange it:

\[ (1 + y + x^2 y) \, dx + (x + x^3) \, dy = 0 \implies (x + x^3) \, dy = -(1 + y + x^2 y) \, dx. \]

Dividing both sides by \(dx\) gives:

\[ \frac{dy}{dx} = -\frac{1 + y + x^2 y}{x + x^3}. \]

### Step 2: Simplify the Right Side
Now we can simplify the right side:

\[ \frac{dy}{dx} = -\frac{1 + y + x^2 y}{x(1 + x^2)}. \]

This can be rewritten as:

\[ \frac{dy}{dx} + \frac{(1 + x^2)}{x(1 + x^2)} y = -\frac{1}{x(1 + x^2)}. \]

### Step 3: Identify P(x) and Q(x)
From the equation, we can identify:

\[ P(x) = \frac{1}{x} \quad \text{and} \quad Q(x) = -\frac{1}{x(1 + x^2)}. \]

### Step 4: Calculate the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx}. \]

Calculating the integral:

\[ \int \frac{1}{x} \, dx = \ln |x| \implies \mu(x) = e^{\ln |x|} = |x|. \]

Since we are typically interested in positive values for \(x\) in this context, we can simplify this to:

\[ \mu(x) = x. \]

### Conclusion
Thus, the integrating factor of the given differential equation is:

\[ \boxed{x}. \]