How to do the following integral: ∫01xa−1log⁡(x)dx where a≥0 ?

Call your integral ${\displaystyle I\left(a\right)}$. Then
${\displaystyle I^{\prime }\left(a\right)={\int }_0^1{x}^{a}dx=\frac{1}{a+1}}$
${\displaystyle I^{\prime }\left(a\right)={\int }_0^1{x}^{a}dx=\frac{1}{a+1}}$
as long as ${\displaystyle a\ge 0}$. Now you need to solve the differential equation
${\displaystyle I^{\prime }\left(a\right)=\frac{1}{a+1}}$
This is a very easy differential equation to solve, and the solution is
${\displaystyle I\left(a\right)=\log (a+1)+C}$
where C is some constant. Now we ask, what is that constant? Notice that
${\displaystyle I\left(0\right)={\int }_0^1\frac{1-1}{\log x}dx=0}$
so we need
${\displaystyle I\left(0\right)=\log \left(1\right)+C=0}$,
or rather ${\displaystyle C=0}$. So we conclude that
${\displaystyle {\int }_0^1\frac{x^a-1}{\log x}dx=\log (a+1)}$,
as you suggested.

We can utilize
${\displaystyle {\int }_0^1{x}^{t}dt=\frac{x-1}{\log \left(x\right)}}$
combined with the substitution ${\displaystyle \mapsto x^{\frac{1}{a}}}$, to get
${\displaystyle {\int }_0^1\frac{x^a-1}{\log \left(x\right)}dx={\int }_0^1\frac{x-1}{\log \left(x\right)}{x}^{\frac{1}{a}-1}dx}$
${\displaystyle ={\int }_0^1{\int }_0^1{x}^{\frac{1}{a}-1}{x}^{t}dtdx}$
${\displaystyle ={\int }_0^1\frac{1}{\frac{1}{a}+t}dt}$
${\displaystyle =\log (\frac{1}{a}+1)-\log \left(\frac{1}{a}\right)}$
${\displaystyle =\log (1+a)}$

We have $x^a-1=e^{a\log (x)}-1$. Hence, the integral is
$I={\int }_0^1\frac{x^a-1}{\log (x)}dx={\int }_0^1(\sum _{k=1}^{\mathrm{\infty }}\frac{a^k{\log }^k(x)}{k!})\frac{dx}{\log (x)}$
$=\sum _{k=1}^{\mathrm{\infty }}\frac{a^k}{k!}{\int }_0^1{\log }^{k-1}(x)dx$
$=\sum _{k=1}^{\mathrm{\infty }}\frac{a^k}{k!}(-1{)}^{k-1}(k-1)!=\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k-1}{a}^k}{k}=\log (1+a)$