Step-by-Step Solution

1. First, rewrite the given differential equation in the standard form: \(\frac{dy}{dx} + P(x)y = Q(x)\). Divide the entire equation by \((1-x^2)\):

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

2. Identify \(P(x)\). In this case, \(P(x) = \frac{x}{1-x^2}\).

3. Calculate the integrating factor (IF) using the formula: \(IF = e^{\int P(x) dx}\). So,

   \(IF = e^{\int \frac{x}{1-x^2} dx}\)

4. To evaluate the integral \(\int \frac{x}{1-x^2} dx\), use substitution. Let \(u = 1-x^2\), then \(du = -2x dx\), so \(x dx = -\frac{1}{2} du\). Therefore,

   \(\int \frac{x}{1-x^2} dx = \int \frac{-\frac{1}{2}}{u} du = -\frac{1}{2} \int \frac{1}{u} du = -\frac{1}{2} \ln|u| = -\frac{1}{2} \ln|1-x^2|\)

5. Since \(-1 \lt x \lt 1\), \(1-x^2\) is always positive, so we can drop the absolute value: \(-\frac{1}{2} \ln(1-x^2)\).

6. Now, substitute this back into the integrating factor formula:

   \(IF = e^{-\frac{1}{2} \ln(1-x^2)} = e^{\ln((1-x^2)^{-\frac{1}{2}})} = (1-x^2)^{-\frac{1}{2}} = \frac{1}{\sqrt{1-x^2}}\)