Fujikawa's calculation of gravitational anomaly

Question

I am trying to reproduce equation 4.23 in this reference Path integral for gauge theories with fermions. It is a computation for gravitational anomaly using the Fujikawa method. I understand, 
\begin{equation}
\frac{1}{32 \pi^{4}} Tr \gamma_{5} \int d^{4} k \left( \frac{1}{2} D^{2}  D^{2}f^{''}(k^{2})
+ \frac{2}{3} k_{\mu} k_{\nu} (D^{2}D^{\mu}D^{\nu} + D^{\mu}D^{2}D^{\nu} + D^{\mu}D^{\nu}D^{2}) f^{'''}(k^{2})+ \frac{2}{3} k_{\mu}k_{\nu}k_{\alpha}k_{\beta} D^{\mu}D^{\nu}D^{\alpha}D^{\beta} f^{iv}(k^{2})  \right)
\end{equation}
Using $k^{2} = \xi$, gives $$\int d^{4} k =  \pi^{2}  \int^{\infty}_{0} \sqrt{g} k^{2} d k^{2} = \pi^{2} \int \sqrt{g}\xi d \xi,$$  and $$ f(\infty) = f^{'}(\infty) = f^{''}(\infty) = ...=0 ,$$ $f(0) =1.$ Using integration by parts $$\int^{\infty}_{0} d \xi \xi f^{''}(\xi) = 1, \int^{\infty}_{0} d \xi \xi^{2} f^{''}(\xi) = -2, \int^{\infty}_{0} d \xi \xi^{3} f^{iv}(\xi) = 6.$$ I finally obtain,
\begin{equation}
Tr \gamma_{5} \frac{\sqrt{g}}{32 \pi^{2}} \left( \frac{1}{2} D^{2} D^{2} - \frac{4}{3} (2 D^{2} D^{2} + D^{\mu}D^{2} D_{\mu}) + 4 (D^{2}D^{2} + D^{\mu} D^{2}D_{\mu} + D^{\mu}D^{\nu}D_{\mu}D_{\nu} \right)
\end{equation} 

I don't see how this finally gives $$\sqrt{g} \frac{1}{192 \pi^{2}} Tr \left( \gamma_{5} [D^{\mu} , D^{\nu}][D_{\mu}D_{\nu}]  \right).$$