You can't directly integrate the ideal gas law because both the temperature and volume vary. You can solve for the temperature in terms of volume and then integrate it, but that just gives you the same $P=\frac{K}{V^\gamma}$.
Your answer should be positive: since the final volume is smaller, work is done on the gas by the environment.
None of your work looks particularly wrong.
I assume by the "density of internal energy" $u$ you mean
$$u = \frac{E_{therm}}{n},$$
where $E_{therm}$ is the thermal energy of the gas, and $n$ is the number of moles.
In this case your equation is true for any ideal gas process. Here is how we can show that this is true.
We begin by considering an isochoric process (V=const). By definition, the amount of heat $Q$ needed to increase the temperature of gas by $\Delta T$ is
$$Q = n C_v \Delta T\tag{1},$$
where $C_v$ is the molar specific heat at constant volume. Since there is no change in volume, the work done is zero. Therefore, from the 1st law of thermodynamics:
$$\Delta E_{therm} = Q + 0 = Q.$$
Combining with Equation (1) we get:
$$\Delta E_{therm} = n C_v \Delta T. \tag{2}$$
Next we apply a key idea from thermodynamics: the change in internal energy $E_{therm}$ of gas is the same for any two processes that results in the same change in temperature $\Delta T$. Therefore, Equation (2) is true for any ideal gas process, and not just the isochoric process.
If we divide both sides by $n$, we get
$$\Delta u = C_v \Delta T,$$
which is what we needed to show.
Best Answer
Based on Adiabatic Equation Wikipedia
$${\displaystyle {\text{(1)}}\qquad dU+\delta W=\delta Q=0,}$$
$${\displaystyle {\text{(2)}}\qquad \delta W=P\,dV.}$$
$${\displaystyle {\text{(3)}}\qquad U=\alpha nRT,}$$
Differentiating Equation (3) and use of the ideal gas law, $PV = nRT$, yields
$${\displaystyle {\text{(4)}}\qquad dU=\alpha nR\,dT=\alpha \,d(PV)=\alpha (P\,dV+V\,dP).}$$