Then we have that
$${\it D \Phi}= \left( {\frac {\partial}{\partial \theta}} -i \theta {\frac {\partial}{\partial t}} \right) \left(
x \left( t \right) +i\theta\,\psi \left( t \right) \right)= i\psi \left( t \right) -i\theta {\frac {d}{dt}}x \left( t \right) $$
and
$${\it D^2 \Phi}= \left( {\frac {\partial}{\partial \theta}} -i \theta {\frac {\partial}{\partial t}} \right) (i\psi \left( t \right) -i\theta {\frac {d}{dt}}x \left( t \right))= -i{\frac {d}{dt}}x \left( t \right) +\theta\,{\frac {d}{dt}}\psi
\left( t \right)$$
Now we have that
$${\it D \Phi}{\it D^2 \Phi}=( i\psi \left( t \right) -i\theta {\frac {d}{dt}}x \left( t \right) )(-i{\frac {d}{dt}}x \left( t \right) +\theta\,{\frac {d}{dt}}\psi \left( t \right))$$
it is to say
$${\it D \Phi}{\it D^2 \Phi}=\left( {\frac {d}{dt}}x \left( t \right) \right) \psi \left( t \right) - \theta( {\frac {d}{dt}}x \left( t \right) ) ^{2}
-i\theta\,\psi \left( t \right) {\frac {d}{dt}}\psi \left( t \right)$$
Finally we have that
$$\int {d\theta {\it D \Phi}{\it D^2 \Phi}}=\int (\left( {\frac {d}{dt}}x \left( t \right) \right) \psi \left( t \right) - \theta( {\frac {d}{dt}}x \left( t \right) ) ^{2}
-i\theta\,\psi \left( t \right) {\frac {d}{dt}}\psi \left( t \right)) {d\theta }$$
it is to say
$$\int {d\theta {\it D \Phi}{\it D^2 \Phi}}={\frac {d}{d\theta}}(\left( {\frac {d}{dt}}x \left( t \right) \right) \psi \left( t \right) - \theta( {\frac {d}{dt}}x \left( t \right) ) ^{2}
-i\theta\,\psi \left( t \right) {\frac {d}{dt}}\psi \left( t \right)) $$
which is reduced to
$$\int {d\theta {\it D \Phi}{\it D^2 \Phi}}=- ( {\frac {d}{dt}}x \left( t \right) ) ^{2}-i\psi \left(
t \right) {\frac {d}{dt}}\psi \left( t \right)$$
From the last equation we deduce that
and then we conclude that there is a missing minus sign in the equation (3.2):