University Calculus: Early Transcendentals (3rd Edition)

$y=\arctan (x) -2 \arcsin x+2$
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 $y=\arctan (x) -2 \arcsin x+2$