Factorization? By Parts?

See this , this is basically defining $\displaystyle F(a)=\int_0^1 \dfrac{x^a-1}{\ln x}\; dx$ and then obtaining $F'(a)=\dfrac{1}{a+1}$ from which one can deduce,

$\displaystyle \begin{aligned} F(3) &=\int_0^3 F'(u)\; du \\ &= \int_0^3 \dfrac{du}{1+u} \\ &= \left[\ln(1+u)\right]_0^3 \\ &= \ln4\end{aligned}$

making the answer $\boxed{4}$

Similar solution with @Aditya Narayan Sharma 's. Refer to Differentiation Under the Integration Sign in Integration Tricks .

$\begin{aligned} I(a) & = \int_0^1 \frac {x^a-1}{\ln x} dx \\ \frac {\partial I(a)}{\partial a} & = \int_0^1 \frac {x^a\ln x}{\ln x} dx = \int_0^1 x^a dx = \frac {x^{a+1}}{a+1} \bigg|_0^1 = \frac 1{a+1} \\ I(a) & = \int \frac 1{a+1} da = \ln (a+1) + \color{#3D99F6} C \quad \quad \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ I(0) & = \ln 1 + C = 0 \\ \implies C & = 0 \\ \implies I(a) & = \ln(a+1) \\ I(3) & = \ln 4 \end{aligned}$

$\implies \exp (I(3)) = \boxed{4}$