## Calculus: Early Transcendentals 8th Edition

$F(x)=4x-3\arctan x + \frac{3\pi}{4} -4$
Note that the antiderivative of a constant is that constant multiplied by the variable. Note that $\frac{d}{dx} \arctan x=\frac{1}{1+x^2}$. Thus $$F(x)=4x-3\arctan x +C$$ Then use the given initial condition to solve for $C.$ $$0=4(1)-3\arctan 1 + C$$ $$0=4-\frac{3\pi}{4} +C$$ $$C=\frac{3\pi}{4}-4$$.