A logarithm in the denominator

Put a parameter in the exponent, differentiate with respect to it, and integrate back.

1. Define $I(a)$ as the integral of $\frac{x^a-1}{\ln x}$ from $0$ to $1$ for $a\ge 0$, so $I(0)=0$ and we want $I(3)$.
2. Differentiating under the integral sign removes the logarithm: $I^{\prime }(a)$ is the integral of $x^a$ from $0$ to $1$, which is $\frac{1}{a+1}$.
3. Integrating back from $0$ gives $I(a)=\ln (a+1)$, so $I(3)=\ln 4$.
4. Since $\ln 4\approx 1.3863$, the integral is $\approx 1.386$.

Answer: $\approx 1.386$