Details of the simple integration by series for $\int_0^1\log(x!)dx$ mentioned above (hopefully yours may be treated analogously, if you wish to try it).

Start from the series of the logarithm of the infinite product for $x!$ $$\log(x!)=- \gamma x + \sum_{k=1}^m\bigg(\frac{x}{k}-\log\big(1+\frac{x}{k}\big)\bigg)+o(1),$$ uniformly on $[0,1]$ as $m\to+\infty$. Integrating on $[0,1]$ we get

$$\int_0^1\log(x!)dx= -\frac{\gamma}{2}+\frac{1}{2}\sum_{k=1}^m \frac{1}{k}+m-\sum_{k=1}^m\Big((k+1)\log(k+1)-k\log k\Big) +\sum_{k=1}^m \log k +o(1)$$ Simplifying the telescopic sum, and by the definition of the Euler-Mascheroni constant, this is

$$= \frac{1}{2}\log m +m -(m+1)\log(m+1)+ \log(m!) +o(1)$$ $$= \frac{1}{2}\log m +m -\Big(m\log m + \log m + 1\Big)+ \log(m!) +o(1)$$ $$=- \frac{1}{2}\log m +m -m\log m -1 + \log(m!) +o(1)$$ Finally by the (logarithm of the) Stirling formula this is exactly $$\frac{1}{2}\log(2\pi)-1+ o(1),$$ ending the computation.

Details of the simple integration by series for $\int_0^1\log(x!)dx$ mentioned above (hopefully yours may be treated analogously, if you wish to try it).

Start from the series of the logarithm of the infinite product for $x!$ $$\log(x!)=- \gamma x + \sum_{k=1}^m\bigg(\frac{x}{k}-\log\big(1+\frac{x}{k}\big)\bigg)+o(1),$$ uniformly on $[0,1]$ as $m\to+\infty$. Integrating on $[0,1]$ we get

$$\int_0^1\log(x!)dx= -\frac{\gamma}{2}+\frac{1}{2}\sum_{k=1}^m \frac{1}{k}+m-\sum_{k=1}^m\Big((k+1)\log(k+1)-k\log k\Big) +\sum_{k=1}^m \log k +o(1)$$ Simplifying the telescopic sum, and by the definition of the Euler-Mascheroni constant, this is

$$= \frac{1}{2}\log m +m -(m+1)\log(m+1)+ \log(m!) +o(1)$$ $$= \frac{1}{2}\log m +m -\Big(m\log m + \log m + 1\Big)+ \log(m!) +o(1)$$ $$=- \frac{1}{2}\log m +m -m\log m -1 + \log(m!) +o(1)$$ Finally by the (logarithm of the) Stirling formula this is exactly $$\frac{1}{2}\log(2\pi)-1+ o(1),$$ ending the computation.

Details of the simple integration by series for $\int_0^1\log(x!)dx$ mentioned in comment in [1].

Start from the series of the logarithm of the infinite product for $x!$ $$\log(x!)=- \gamma x + \sum_{k=1}^m\bigg(\frac{x}{k}-\log\big(1+\frac{x}{k}\big)\bigg)+o(1),$$ uniformly on $[0,1]$ as $m\to+\infty$. Integrating on $[0,1]$ we get

$$\int_0^1\log(x!)dx= -\frac{\gamma}{2}+\frac{1}{2}\sum_{k=1}^m \frac{1}{k}+m-\sum_{k=1}^m\Big((k+1)\log(k+1)-k\log k\Big) +\sum_{k=1}^m \log k +o(1)$$ Simplifying the telescopic sum, and by the definition of the Euler-Mascheroni constant, this is

$$= \frac{1}{2}\log m +m -(m+1)\log(m+1)+ \log(m!) +o(1)$$ $$= \frac{1}{2}\log m +m -\Big(m\log m + \log m + 1\Big)+ \log(m!) +o(1)$$ $$=- \frac{1}{2}\log m +m -m\log m -1 + \log(m!) +o(1)$$ Finally by the (logarithm of the) Stirling formula this is exactly $$\frac{1}{2}\log(2\pi)-1+ o(1),$$ ending the computation.

Details of the simple integration by series for $\int_0^1\log(x!)dx$ mentioned above (hopefully yours may be treated analogously, if you wish to try it).

Start from the series of the logarithm of the infinite product for $x!$ $$\log(x!)=- \gamma x + \sum_{k=1}^m\bigg(\frac{x}{k}-\log\big(1+\frac{x}{k}\big)\bigg)+o(1),$$ uniformly on $[0,1]$ as $m\to+\infty$. Integrating on $[0,1]$ we get

$$\int_0^1\log(x!)dx= -\frac{\gamma}{2}+\frac{1}{2}\sum_{k=1}^m \frac{1}{k}+m-\sum_{k=1}^m\Big((k+1)\log(k+1)-k\log k\Big) +\sum_{k=1}^m \log k +o(1)$$ Simplifying the telescopic sum, and by the definition of the Euler-Mascheroni constant, this is

$$= \frac{1}{2}\log m +m -(m+1)\log(m+1)+ \log(m!) +o(1)$$ $$= \frac{1}{2}\log m +m -\Big(m\log m + \log m + 1\Big)+ \log(m!) +o(1)$$ $$=- \frac{1}{2}\log m +m -m\log m -1 + \log(m!) +o(1)$$ Finally by the (logarithm of the) Stirling formula this is exactly $$\frac{1}{2}\log(2\pi)-1+ o(1),$$ ending the computation.