Quick questions on Derivatives and integrals of inverse trigonometric functions

The skill examined here is recognising which standard form an integrand matches. A square root of $a^2 - x^2$ in the denominator signals an $\arcsin$ result; a sum $a^2 + x^2$ in the denominator (no root) signals an $\arctan$ result. Always identify $a$ first: in $\frac{1}{9 + x^2}$ you have $a^2 = 9$, so $a = 3$ and the constant in front of $\arctan$ is $\frac{1}{a} = \frac{1}{3}$. A common slip is to read $a^2$ itself as $a$, which gives the wrong constant.

These integrals are nothing more than the derivative results read backwards, so the fastest way to recall them under exam pressure is to recall the three derivatives and reverse each one. If $\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}$, then $\int \frac{1}{1 + x^2}\,dx = \arctan x + C$. Keeping the derivative-integral pairs together in your memory halves what you need to learn and guards against the sign error on the $\arccos$ form, which never appears as a standard integral precisely because it duplicates the $\arcsin$ result up to sign.

$\arcsin$ derivative is undefined at $x = \pm 1$, where the tangent is vertical.

$\int \frac{dx}{4 + x^2}$ has $a = 2$ (not $4$), so the constant in front of $\arctan$ is $\frac{1}{2}$ (not $\frac{1}{4}$).