Fubini saves you

$\begin{aligned}\text{Let }I(a)&=\int_0^1\frac{t^a-1}{\ln t}dt\\ \therefore\frac d{da}I(a)&=\int_0^1\frac d{da}\frac{t^a-1}{\ln t}dt&\text{using differentiation under the integral sign}\\ &=\int_0^1\frac{t^a\ln t}{\ln t}dt\\ &=\int_0^1 t^a\,dt\\ &=\left[\frac{t^{a+1}}{a+1}\right]_0^1\\ &=\frac1{a+1}\\ \therefore I(a)&=\int\frac1{a+1}da\\ &=\ln(a+1)+C\\ \text{now, }I(0)&=\int_0^1\frac{t^0-1}{\ln t}dt=\int_0^1\frac{1-1}{\ln t}dt=0\\ \therefore\ln(0+1)+C&=0\\ \therefore C&=0\\ \therefore I(a)&=\ln(a+1)\\ \therefore \int_0^1\frac{t^3-1}{\ln t}&=I(3)=\ln4\end{aligned}$