Solution:

$\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\dots (1)$
General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=$ Qis given by, $y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c$ Where, integrating factor,
$\begin{array}{c}
I . F .=e \int p d x \\
\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0 \\
\left(1+y^{2}\right) d x=-\left(x-e^{-\tan ^{-1} y}\right) d y \\
\frac{d x}{d y}=\frac{\left(e^{-\tan ^{-1} y}-x\right)}{\left(1+y^{2}\right)}
\end{array}$
$\frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{-\tan ^{-1} y}}{1+y^{2}} \ldots \ldots \ldots(1)$
In equation (1)
$\mathrm{P}=1 /\left(1+y^{2}\right) \text { and } \mathrm{Q}=\frac{e^{-\tan ^{-1} y}}{1+y^{2}}$
Therefore, integrating factor is
$e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}$
$\text { I.F. }=e \int^{p d x}$
General solution is
$\begin{array}{l}
y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c \\
x \cdot e^{-\tan ^{-1} y}=\int \frac{e^{-\tan ^{-1} y}}{1+y^{2}} \cdot e^{\tan ^{-1} y} d y+C \\
x\left(e^{\tan ^{-1} y}\right)=\tan ^{-1} y+c
\end{array}$
Putting $x=0$ and $y=0$
$\mathrm{c}=0$
Therefore, general solution is
$\text { x }\left(e^{\tan ^{-1} y}\right)=\tan ^{-1} y$