$dy/dx = (9x + 4y + 1)^2$

$dy = (9x + 4y + 1)^2~dx$

Let

$z = 9x + 4y + 1$
$dz = 9~dx + 4~dy$

$dy = \frac{1}{4}(dz - 9~dx)$

$\frac{1}{4}(dz - 9~dx) = z^2~dx$

$dz - 9~dx = 4z^2~dx$

$dz = 4z^2~dx + 9~dx$

$dz = (4z^2 + 9)~dx$

$\dfrac{dz}{4z^2 + 9} = dx$

$\dfrac{dz}{(2z)^2 + 3^2} = dx$

$\displaystyle \dfrac{1}{2} \int \dfrac{2dz}{(2z)^2 + 3^2} = \int dx$

$\dfrac{1}{2} \left[ \dfrac{1}{3} \arctan \left( \dfrac{2z}{3} \right) \right] = x + \frac{1}{6}c$

$\frac{1}{6} \arctan (\frac{2}{3}z) = x + \frac{1}{6}c$

$\arctan (\frac{2}{3}z) = 6x + c$

$\frac{2}{3}z = \tan (6x + c)$

$2z = 3\tan (6x + c)$

But $z = 9x + 4y + 1$

$2(9x + 4y + 1) = 3\tan (6x + c)$

$3\tan (6x + c) = 2(9x + 4y + 1)$ *answer*