First Order Non-Linear Equations please ; How can solve ODE.
$$(1+y^2)+(x-e^{tan^{-1}y})\frac {dy}{dx}=0$$
These my attempt
$$ let\space u=tan^{-1}y      $$
$$  y= tanu$$
$$\frac{dy}{dx}= sec^{2}u \frac{du}{dx}$$
$$(1+tan^{2}u)+(x-e^{u})\sec^{2}u\frac{du}{dx}=0$$
$$(sec^{2}u)+(x-e^{u})\sec^{2}u\frac{du}{dx}=0$$
$$(x-e^{u})\frac{du}{dx}=-1$$
This non-Linear ODE how can solved it?? please

using a substitution as follows
1) $ u = \ln(v)$ we find
$$
\left(x-v\right)\frac{v'}{v} = -1
$$
re-arranging gives
$$
vv' -v - xv' = \frac{d}{dx}\frac{v^{2}}{2}-\frac{d}{dx}(xv) = 0
$$
this yields
$$
\frac{v^{2}}{2} -xv + C = 0
$$
or
$$
v^{2} - 2xv + C = 0
$$
the solution is 
$$
v = x \pm \sqrt{x^{2}-C},
u = \ln\left[x \pm \sqrt{x^{2}-C}\right]
$$
thus
$$
y = \tan\left[\ln\left(x \pm \sqrt{x^{2}-C}\right)\right]
$$
To check that this is indeed the set of solutions, we take the positive root solution and take the derivative,
$$
\frac{dy}{dx} = \left(1+y^{2}\right)\left(\frac{1+\frac{x}{\sqrt{x^{2}-C}}}{x + \sqrt{x^{2}-C}}\right) = \frac{\left(1+y^{2}\right)}{\sqrt{x^{2}-C}}
$$
inserting into the original equation 
$$
\left(1+y^{2}\right) + \left(x - x-\sqrt{x^2 -C}\right)\frac{\left(1+y^{2}\right)}{\sqrt{x^{2}-C}} = \left(1+y^{2}\right)  - \left(1+y^{2}\right) = 0.
$$
