What are the usual ways to follow in order to solve the integrals given below?...

As an addendum of sorts to the previous answers, there is the identity
${\displaystyle \log G(z+1)=\frac{z}{2}\log \left(2\pi \right)-\frac{z(z+1)}{2}+z\log \mathrm{\Gamma }(z+1)-z(\log z-1)-{\int }_0^z\log \mathrm{\Gamma }\left(t\right)dt}$
where ${\displaystyle \log G\left(z\right)}$ is the logarithm of the Barnes function (double gamma function) ${\displaystyle G\left(z\right)}$, the function that satisfies the functional equation ${\displaystyle G(z+1)=\mathrm{\Gamma }\left(z\right)G\left(z\right)}$. (Barnes proved this identity in his paper, where he introduced the function now named after him.) For n an integer, ${\displaystyle G\left(n\right)}$ can be expressed as
${\displaystyle G\left(n\right)=\prod _{k=1}^{n-2}k!}$
Thus, to evaluate ${\displaystyle {\int }_0^1\log \mathrm{\Gamma }\left(t\right)dt}$, we have
${\displaystyle \log G\left(2\right)=\frac{1}{2}\log \left(2\pi \right)-1+\log \mathrm{\Gamma }\left(2\right)-(\log 1-1)-{\int }_0^1\log \mathrm{\Gamma }\left(t\right)dt}$
${\displaystyle 0=\frac{1}{2}\log \left(2\pi \right)-{\int }_0^1\log \mathrm{\Gamma }\left(t\right)dt}$
and you obtain the same solution as Andrew.
For the integral ${\displaystyle {\int }_0^{1}t\log \mathrm{\Gamma }\left(t\right)dt}$, integration by parts and taking appropriate limits yields the identity
${\displaystyle {\int }_0^{1}t\log \mathrm{\Gamma }\left(t\right)dt=-\frac{1}{2}{\int }_0^1{t}^2\psi \left(t\right)dt}$
Now, Victor Adamchik, in a paper on negative-order polygamma functions (the same sort of functions that appear in Argon's answer), gives the identity
${\displaystyle {\int }_0^z{x}^n\psi \left(x\right)dx=}$
${\displaystyle ={(-1)}^n(\frac{B_{n+1}{H}_n}{n+1}-{\zeta }^{\prime }(-n))+\sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right)z^{n-k}(\zeta (-k,z)-\frac{B_{k+1}\left(z\right)H_k}{k+1}}$
where ${\displaystyle B_n}$ and