Ta có: $x=x(2+\tan^2x) \Leftrightarrow x=0$
Diện tích hình phảng cần tìm là:
$S=\int\limits_0^{\frac{\pi}{4}}(x(2+\tan^2x)-x)dx$
$=\int\limits_0^{\frac{\pi}{4}}x(1+\tan^2x)dx$
$=\int\limits_0^{\frac{\pi}{4}}xd(\tan x)$
$=x\tan x \left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.-\int\limits_0^{\frac{\pi}{4}}\tan xdx$
$=\dfrac{\pi}{4}+\int\limits_0^{\frac{\pi}{4}}\dfrac{d(\cos x)}{\cos x}$
$=\dfrac{\pi}{4}+\ln(\cos x)\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.$
$=\dfrac{\pi}{4}-\dfrac{\ln 2}{2}$