To find the indefinite integral of $\frac{1 - \cos(x)}{(x - \sin(x))^2}$, we can use integration techniques. However, this integral is quite complex, and direct integration is not straightforward. Let's start by attempting to simplify the expression and see if we can find a more manageable form for integration.

Consider the substitution $u = x - \sin(x)$. Then, we have:

$du = \left(1 - \cos(x)\right)dx$

Now, we can rewrite the integral in terms of $u$:

$\int \frac{1 - \cos(x)}{(x - \sin(x))^2} \, dx = \int \frac{1 - \cos(x)}{u^2} \, dx$

Since $du = (1 - \cos(x)) dx$, we get $dx = \frac{du}{1 - \cos(x)}$.

Substituting this into the integral, we get:

$\int \frac{1 - \cos(x)}{u^2} \cdot \frac{du}{1 - \cos(x)} = \int \frac{du}{u^2}$

The integral of $\frac{1}{u^2}$ is straightforward:

$\int \frac{1}{u^2} \, du = -\frac{1}{u} + C$

Substituting back $u = x - \sin(x)$, we get:

$-\frac{1}{x - \sin(x)} + C$

So, the indefinite integral of $\frac{1 - \cos(x)}{(x - \sin(x))^2}$ is:

$\boxed{-\frac{1}{x - \sin(x)} + C}$

Would you like further details or have any questions?

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Tip: When dealing with complex integrals, substitution can often simplify the process by transforming the integral into a more manageable form.