Answer
$y(x)=tan^{-1}(1+Ce^{-\sqrt 1+ x})$
Work Step by Step
We are given:
$$\sec^2y\frac{dy}{dx}+\frac{1}{2\sqrt 1+x}\tan y=\frac{1}{2\sqrt 1+x}$$
with $p(x)=\frac{1}{2\sqrt 1+x}$
$q(x)=\frac{1}{2\sqrt 1+x}$
$\tan y= f( y) \rightarrow \sec^2y=f'(y)$
Let $t=\tan y \rightarrow t'=sec^2yy'$
Substituting:
$$\frac{du}{dx}+\frac{1}{2\sqrt 1+x}t=\frac{1}{2\sqrt 1+x}$$
The integrating factor is:
$$I(x)=e^{\frac{1}{2}\int \frac{1}{\sqrt 1+x}dx}=e^{\sqrt 1+x}$$
Hence,
$$\frac{d}{dx}(e^{\sqrt 1+x}t)=\frac{e^{\sqrt 1+x}}{2\sqrt 1+x}$$
Integrating both sides:
$$e^{\sqrt 1+x}t=e^{\sqrt 1+x}+C$$
$C$ is a constant of integration.
$$t=1+\frac{C}{e^{\sqrt 1+x}}$$
Since $t= \tan y \rightarrow y=\tan^{-1} y$
and $\tan x$ is an invertible function
The solution is:
$$y(x)=tan^{-1}(1+Ce^{-\sqrt 1+ x})$$
