The integral I = ∫ 2sin x (3 + 2x)dx simplifies to (where, C is the...

The given integral can be rewritten as,

$I = ∫ \dfrac{sin x + cos x}{3 + \sin 2x}dx + ∫ \dfrac{sin x - cos x}{3 + \sin 2x}dx$
Let $sin x - cos x = t$ and $sin x + cos x = u$ in ${1}^{st}$ and ${2}^{nd}$ integral respectively.
$∴ I = ∫ \dfrac{dt}{4 - t^{2}} - ∫ \dfrac{du}{2 + u^{2}}$
$= \dfrac{1}{4}\ln \left|\dfrac{2 + t}{2 - t}\right| - \dfrac{1}{2}\tan^{-1} \left(\dfrac{u}{2}\right) + C$
Putting values of $t$ and $u$, we get,

$I = \dfrac{1}{4}\ln \left|\dfrac{2 + \sin x - \cos x}{2 - \sin x + \cos x}\right| - \dfrac{1}{2}\tan^{-1} \left(\dfrac{\sin x + \cos x}{2}\right) + C$