Answer
$y=\arctan (x) -2 \arcsin x+2$
Work Step by Step
We need to integrate:
$y=\int \dfrac{dx}{1+x^2} \int \dfrac{2dx}{\sqrt{1-x^2}}$
We use the differential trigonometric formula:
$\int \dfrac{dx}{\sqrt{x^2-a^2}}=\arcsin (x/a)+C$ and $\int \dfrac{1}{x^2+a^2}=\arctan (x/a)+C$
$y=\arctan (x) -2 \arcsin x+c$
Take initial conditions: $y(0)=2$
Then $y(0)=\arctan (0) -2 \arcsin 0+c =2 \implies c=2$
Thus, we have $y=\arctan (x) -2 \arcsin x+2$
