Let y=y(x) be the solution curve of the differential equation $$(1+x^{2})dy+(y-\tan^{-1}x)dx=0,y(0)=1$$. Then the value of y (1) is :

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

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

IF = $$e^{\int \frac{dx}{1+x^2}} = e^{\tan^{-1}x}$$.

$$ye^{\tan^{-1}x} = \int \frac{\tan^{-1}x}{1+x^2}e^{\tan^{-1}x}dx$$

Let $$t = \tan^{-1}x$$: $$= \int te^t dt = te^t - e^t + C = e^t(t-1) + C$$.

$$ye^{\tan^{-1}x} = e^{\tan^{-1}x}(\tan^{-1}x - 1) + C$$

At $$x = 0$$: $$1 \cdot e^0 = e^0(0-1) + C \Rightarrow 1 = -1 + C \Rightarrow C = 2$$.

$$y = \tan^{-1}x - 1 + 2e^{-\tan^{-1}x}$$

At $$x = 1$$: $$y(1) = \frac{\pi}{4} - 1 + 2e^{-\pi/4} = \frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1$$.

The answer is Option 2: $$\frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1$$.