Given that $y = \sec x$, show that $\frac{dy}{dx} = \sec x \tan x$. Hence show that $y^4 - y^2 = \left(\frac{dy}{dx}\right)^2$. Given that $\frac{6x + 1}{2x + 1}$ can be written in the form $A + \frac{B}{2x + 1}$, find $A$ and $B$. Hence, or otherwise $\int \frac{6x + 1}{2x + 1} dx$.

1. Express $\sec x$ in terms of $\cos x$ and differentiate using the chain rule.

$y = \sec x = (\cos x)^{-1}$

$\frac{dy}{dx} = -1(\cos x)^{-2}(-\sin x)$

2. Simplify the derivative to show the required result.

$\frac{dy}{dx} = \frac{\sin x}{\cos^2 x} = \left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) = \sec x \tan x$

3. Express $\left(\frac{dy}{dx}\right)^2$ in terms of $\sec x$.

$\left(\frac{dy}{dx}\right)^2 = (\sec x \tan x)^2 = \sec^2 x \tan^2 x$

4. Use the Pythagorean identity to show the final relationship.

Using $\tan^2 x = \sec^2 x - 1$:

$\left(\frac{dy}{dx}\right)^2 = \sec^2 x (\sec^2 x - 1) = \sec^4 x - \sec^2 x$

Since $y = \sec x$, we have:

$\left(\frac{dy}{dx}\right)^2 = y^4 - y^2$ (Shown)

1. Express the rational function in the given form to find A and B

$\frac{6x + 1}{2x + 1} = \frac{A(2x + 1) + B}{2x + 1}$
$6x + 1 = 2Ax + A + B$
Comparing coefficients of $x$:
$2A = 6 \implies A = 3$
Comparing constant terms:
$A + B = 1 \implies 3 + B = 1 \implies B = -2$

2. Rewrite the integral using the values of A and B

$\int \frac{6x + 1}{2x + 1} dx = \int \left(3 - \frac{2}{2x + 1}\right) dx$

3. Evaluate the integral

$= 3x - \frac{2}{2}\ln(2x + 1) + C$
$= 3x - \ln(2x + 1) + C$