Why is $\int_0^1 \frac{t^x - 1}{\log(t)} dt$ finite for $x \in [0,1]$? I want to show that the following integral
$$\int_0^1 \frac{t^x - 1}{\log(t)} dt$$ 
is finite for all $x \in [0,1]$. But I struggle to come up with a good argument for that. Thanks in advance :)

IF we differentiate under the integral sign we have
\begin{array}
$ I(x)&=&\displaystyle\int_0^1\frac{t^x-1}{\log t}dt\\
I'(x)&=&\displaystyle\int_0^1\frac{t^x\log t}{\log t}dt\\
&=&\displaystyle\int_0^1 t^x dt\\
&=&\displaystyle\frac{1}{1+x}, \quad x\neq -1
\end{array}
Next integrating we have $I(x)=\log(1+x)+c$. Since $I(0)=0$, $c=0$, we are done.
