Ta có:
$\int\limits_0^{\frac{\pi}{4}}x(1+\sin2x)dx$
$=\int\limits_0^{\frac{\pi}{4}}xdx+\int\limits_0^{\frac{\pi}{4}}x\sin2xdx$
$=\dfrac{x^2}{2}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.+\int\limits_0^{\frac{\pi}{4}}xd(\sin^2x)$
$=\dfrac{\pi^2}{32}+x\sin^2x\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.-\int\limits_0^{\frac{\pi}{4}}\sin^2xdx$
$=\dfrac{\pi^2}{32}+\dfrac{\pi}{8}-\dfrac{1}{2}\int\limits_0^{\frac{\pi}{4}}(1-\cos2x)dx$
$=\dfrac{\pi^2}{32}+\dfrac{\pi}{8}-\dfrac{2x-\sin2x}{4}\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.$
$=\dfrac{\pi^2}{32}+\dfrac{1}{4}$