Ta có
$I = \dfrac{1}{\sqrt{2}} \displaystyle \int \dfrac{\sin x - \cos x}{[\sin(2x) + 1] + 1 + 2(\sin x + \cos x)}dx$
$= -\dfrac{1}{\sqrt{2}} \displaystyle \int \dfrac{d(\sin x + \cos x)}{(\sin x + \cos x)^2 + 2(\sin x + \cos x) + 1}$
Đặt $u = \sin x + \cos x$. Khi đó
$I = -\dfrac{1}{\sqrt{2}} \displaystyle \int \dfrac{du}{u^2 + 2u + 1}$
$= -\dfrac{1}{\sqrt{2}} \displaystyle \int \dfrac{du}{(u+1)^2}$
$= \dfrac{1}{\sqrt{2}}. \dfrac{1}{u + 1} + c$
$= \dfrac{1}{(\sin x + \cos x + 1)\sqrt{2}} + c$
