## Thomas' Calculus 13th Edition

$1.5574$
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 $y(1)=\tan x =\tan (1) \approx 1.5574$