We are given the differential equation:

$$(1+x^2) dy + (1-\tan^{-1} x) dx = 0$$

We can rewrite this as:

$$dy = -\frac{1-\tan^{-1} x}{1+x^2} dx$$

Integrating both sides, we get:

$$\int dy = -\int \frac{1-\tan^{-1} x}{1+x^2} dx$$ $$y = -\int \frac{1}{1+x^2} dx + \int \frac{\tan^{-1} x}{1+x^2} dx$$ $$y = -\tan^{-1} x + \int \tan^{-1} x \cdot \frac{1}{1+x^2} dx$$

Let $u = \tan^{-1} x$, then $du = \frac{1}{1+x^2} dx$. So the integral becomes:

$$\int u du = \frac{u^2}{2} + C = \frac{(\tan^{-1} x)^2}{2} + C$$

Therefore, the general solution is:

$$y = -\tan^{-1} x + \frac{(\tan^{-1} x)^2}{2} + C$$

We are given that $y(0) = 1$. Plugging in $x=0$ and $y=1$, we get:

$$1 = -\tan^{-1} (0) + \frac{(\tan^{-1} (0))^2}{2} + C$$ $$1 = -0 + \frac{0^2}{2} + C$$ $$C = 1$$

So the particular solution is:

$$y(x) = -\tan^{-1} x + \frac{(\tan^{-1} x)^2}{2} + 1$$

Now we need to find $y(1)$. Plugging in $x=1$, we get:

$$y(1) = -\tan^{-1} (1) + \frac{(\tan^{-1} (1))^2}{2} + 1$$ $$y(1) = -\frac{\pi}{4} + \frac{(\frac{\pi}{4})^2}{2} + 1$$ $$y(1) = -\frac{\pi}{4} + \frac{\frac{\pi^2}{16}}{2} + 1$$ $$y(1) = -\frac{\pi}{4} + \frac{\pi^2}{32} + 1$$ $$y(1) = \frac{\pi^2}{32} - \frac{\pi}{4} + 1$$