Answer
$1.5574$
Work Step by Step
Here, $\int \dfrac{1}{1+y^2}dy=\int dx$
and $\arctan (y)=x+c$
Further, apply the initial conditions, to calculate the value of $c$
$\arctan (0)=0+c$
or, $c=0$
Now, the particular solution as follows:
$\arctan (y)=x$
This can be re-arranged as: $y=\tan x$
and
$y(1)=\tan x =\tan (1) \approx 1.5574$