I'll look at the first one.
∫sec²(4x+1)dx
The first thing that you need to know is that the integral of tan x is sec2x
My strategy is to work backwards from there.
$$\\\frac{d}{dx}tan(4x+1)=4sec^2(4x+1)$$
Mmm I need to get rid of that pesky 4 so
$$\\\frac{d}{dx}\frac{tan(4x+1)}{4}=\frac{4sec^2(4x+1)}{4}=sec^2(4x+1)\\\\$$
so
$$\int{sec^2(4x+1)dx=\frac{tan(4x+1)}{4}+c$$
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