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(A) The given differential equation is $(1+x^{2})dy + (y-	an^{-1}x)dx = 0$.Dividing by $(1+x^{2})dx$, we get $\frac{dy}{dx} + \frac{y}{1+x^{2}} = \fr

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$\frac{2}{e^{\pi/4}}+\frac{\pi}{4}-1$

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(A) The given differential equation is $(1+x^{2})dy + (y-	an^{-1}x)dx = 0$.Dividing by $(1+x^{2})dx$, we get $\frac{dy}{dx} + \frac{y}{1+x^{2}} = \frac{	an^{-1}x}{1+x^{2}}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{1+x^{2}}$ and $Q = \frac{	an^{-1}x}{1+x^{2}}$.The integrating factor $I.F. = e^{\int P dx} = e^{\int \frac{1}{1+x^{2}} dx} = e^{	an^{-1}x}$.The general solution is $y \cdot I.F. = \int Q \cdot I.F. dx + c$.$y \cdot e^{	an^{-1}x} = \int \frac{	an^{-1}x}{1+x^{2}} e^{	an^{-1}x} dx + c$.Let $t = 	an^{-1}x$, then $dt = \frac{1}{1+x^{2}} dx$.The integral becomes $\int t e^{t} dt = t e^{t} - e^{t} + c$.So, $y \cdot e^{	an^{-1}x} = (	an^{-1}x) e^{	an^{-1}x} - e^{	an^{-1}x} + c$.Given $y(0) = 1$, we have $1 \cdot e^{0} = (0) e^{0} - e^{0} + c \Rightarrow 1 = -1 + c \Rightarrow c = 2$.Thus, $y \cdot e^{	an^{-1}x} = (	an^{-1}x - 1) e^{	an^{-1}x} + 2$.Dividing by $e^{	an^{-1}x}$, we get $y = 	an^{-1}x - 1 + 2e^{-	an^{-1}x}$.At $x = 1$, $y(1) = 	an^{-1}(1) - 1 + 2e^{-	an^{-1}(1)} = \frac{\pi}{4} - 1 + \frac{2}{e^{\pi/4}}$.

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

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