The value of $\int_1^a \frac{x^a - 1}{\log x} dx$ is:
Step-by-Step Solution
Key Concept: Differentiate under the integral sign (Leibniz rule) to convert a difficult integral into an elementary one, then integrate back and apply initial conditions.
Given $I(a) = \int_0^1 \frac{x^a - 1}{\log x}dx$, differentiate with respect to $a$: $\frac{dI}{da} = \int_0^1 \frac{x^a \log x}{\log x}dx = \int_0^1 x^a dx = \frac{x^{a+1}}{a+1}\Big|_0^1 = \frac{1}{a+1}$. Integrating: $I(a) = \log(a+1) + c$. Using $I(0) = 0$, we get $c = 0$, so $I(a) = \log(a+1)$.
Correct Answer: 2