Calculate the following indefinite integral Calculate: $$\int \frac{(2x^{2}+x+\frac{1}{2})\cos2x+(6x^{2}-7x+\frac{13}{2})\sin2x}{\sqrt{(x^{2}-x+1)^{3}}}dx.$$
I have no idea how to start it.

Let $u = x^2-x+1$, we have $u' = 2x-1$. Notice
$$\begin{align}
2x^2+x+\frac12 &= 2(x^2-x+1) + \frac32 (2x-1) = 2u + \frac32 u'\\
  6x^2-7x+\frac{13}{2} &= 6(x^2-x+1) - \frac12(2x-1) = 6u - \frac12 u'
\end{align}
$$
The integrand of the integral can be rewritten as
$$\begin{align}
 & \frac{(2u + \frac32 u')\cos(2x) + (6u - \frac12u')\sin(2x)}{u^{3/2}}\\
= & \left(-\frac12 \frac{u'}{u^{3/2}}\right)(\sin(2x)-3\cos(2x)) 
+  \frac{1}{u^{1/2}}(\sin(2x)-3\cos(2x))'\\
= & \left(\frac{\sin(2x)-3\cos(2x)}{u^{1/2}}\right)'
\end{align}$$
This implies the indefinite integral equals to
$$\frac{\sin(2x)-3\cos(2x)}{\sqrt{x^2-x+1}} + \text{ constant }$$
