I need help understanding the following integrals. Basically, I need help demonstrating them.
$$ \int \frac{du}{a²+u^2} = \frac{1}{a} \tan^{-1}\frac{u}{a} + C$$
$$ \int \frac{du}{\sqrt{a²-u^2}} = \sin^{-1}\frac{u}{a} + C$$
$$ \int \frac{du}{u(\sqrt{u²+a^2})} = \frac{1}{a} \sec^{-1}\mid\frac{u}{a}\mid + C$$
The way you find the derivatives of the inverse trig functions involves implicit differentiation and constructing right triangles. Assuming you know how to do this, you get
$$\frac{d}{du}\arctan(u)=\frac{1}{u^2+1}$$
Fundamental Thm of Calculus tells us this is equivalent to
$$\int\frac{du}{u^2+1}=\arctan(u)+C$$
To obtain the formula for arctangent you've shown simply requires some manipulation.
$$\int\frac{du}{u^2+a^2}=\int\frac{1}{a^2}\frac{du}{(u/a)^2+1}=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C$$
Try using this approach on the other functions listed.