Finding the integral $\int_0^1 \frac{x^a - 1}{\log x} dx$

Call your integral $I(a)$. Then $$ I'(a) = \int_0^1 x^a dx = \frac{1}{a+1} $$ as long as $a \geq 0$. Now you need to solve the differential equation $$ I'(a) = \frac{1}{a + 1}.$$ This is a very easy differential equation to solve, and the solution is $$ I(a) = \log(a+1) + C $$ where $C$ is some constant. Now we ask, what is that constant? Notice that $$ I(0) = \int_0^1 \frac{1 - 1}{\log x} dx = 0,$$ so we need $$ I(0) = \log(1) + C = 0,$$ or rather $C = 0$. So we conclude that $$ \int_0^1 \frac{x^a - 1}{\log x} dx = \log(a + 1), $$ as you suggested. $\diamondsuit$

We can utilize $$ \int_0^1x^t\,\mathrm{d}t=\frac{x-1}{\log(x)} $$ combined with the substitution $x\mapsto x^{1/a}$, to get $$ \begin{align} \int_0^1\frac{x^a-1}{\log(x)}\,\mathrm{d}x &=\int_0^1\frac{x-1}{\log(x)}x^{\frac1a-1}\,\mathrm{d}x\\ &=\int_0^1\int_0^1x^{\frac1a-1}x^t\,\mathrm{d}t\,\mathrm{d}x\\ &=\int_0^1\int_0^1x^{\frac1a-1}x^t\,\mathrm{d}x\,\mathrm{d}t\\ &=\int_0^1\frac1{\frac1a+t}\,\mathrm{d}t\\ &=\log\left(\frac1a+1\right)-\log\left(\frac1a\right)\\[9pt] &=\log(1+a) \end{align} $$