Calculating traces for triangle diagrams with massless fermions
I am following <PERSON> Quantum Field Theory textbook. In particular, I am looking at triangle diagrams with massless fermions. On pg. 623 - 624 <PERSON> attempts to calculate $q_\mu^1 M_{5}^{\alpha\mu\nu}$ which should be zero by the Ward Identity of QED. 346
This gives (Eq. 30.29),
\begin{multline} q_\mu^1 M_{5}^{\alpha\mu\nu} = \int \frac{d^4 k}{(2\pi)^2} \left[ \frac{\text{Tr}[ \gamma^\nu(\not! k + \not!q_2)\gamma^\alpha\gamma^5(\not! k - \not!q_1) ]}{(k - q_1)^2 (k + q_2)^2} - \frac{\text{Tr}[ \not!k \gamma^\nu(\not! k + \not! q_2)\gamma^\alpha\gamma^5 ]}{k^2 (k + q_2)^2} \right.\[0.25cm] \left. + \frac{ \text{Tr}[\gamma^\nu\not!k \gamma^\alpha\gamma^5(\not! k - \not! q_2)] }{k^2(k - q_2)^2} - \frac{\text{Tr}[ \gamma^\nu(\not! k + \not! q_1)\gamma^\alpha\gamma^5(\not! k - \not! q_2) ]}{(k + q_1)^2(k - q_2)^2} \right] \end{multline}
<PERSON> says that after completing the traces, the result is (Eq. 30.30) \begin{equation} q_\mu^1 M_{5}^{\alpha\mu\nu} = - 4i\epsilon^{\alpha\nu\rho\sigma} \int \frac{d^4 k}{(2\pi)^2} \left[ \frac{(k - q_1)^\rho(k + q_2)^\sigma}{(k - q_1)^2(k + q_2)^2} - \frac{(k - q_2)^\rho(k + q_1)^\sigma}{(k - q_2)^2(k + q_1)^2} \right] \end{equation}
This leads me to believe that $$ I = \int \frac{d^4 k}{(2\pi)^2} \left[ - \frac{\text{Tr}[ \not!k \gamma^\nu(\not! k + \not! q_2)\gamma^\alpha\gamma^5 ]}{k^2 (k + q_2)^2} + \frac{ \text{Tr}[\gamma^\nu\not!k \gamma^\alpha\gamma^5(\not! k - \not! q_2)] }{k^2(k - q_2)^2} \right] = 0 $$ which I am having difficulty proving. So far, I have evaluated the traces as follows \begin{align} \text{Tr}[ \not!k \gamma^\nu(\not! k + \not! q_2)\gamma^\alpha\gamma^5 ] &= \text{Tr}[ \gamma^\alpha\gamma^5 \not!k \gamma^\nu(\not! k + \not! q_2) ]\[0.25cm] &= -\text{Tr}[\gamma^5\gamma^\alpha\gamma^\rho\gamma^\nu\gamma^\sigma ]k_\rho(k + q_2)\sigma\[0.25cm] &= -4i\epsilon^{\alpha\rho\nu\sigma} k\rho(k + q_2)\sigma\[0. 262