The value of the integral int_0^1 frac{x^b - | vedclass

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(D) Let $I(b) = \int_0^1 \frac{x^b - 1}{\log x} \, dx$. Using Feynman's technique of differentiation under the integral sign,we differentiate $I(b)$ w

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(D) Let $I(b) = \int_0^1 \frac{x^b - 1}{\log x} \, dx$. Using Feynman's technique of differentiation under the integral sign,we differentiate $I(b)$ with respect to $b$: $I'(b) = \frac{d}{db} \int_0^1 \frac{x^b - 1}{\log x} \, dx = \int_0^1 \frac{\partial}{\partial b} \left( \frac{x^b - 1}{\log x} \right) \, dx$. Since $\frac{\partial}{\partial b} (x^b) = x^b \log x$,we have: $I'(b) = \int_0^1 \frac{x^b \log x}{\log x} \, dx = \int_0^1 x^b \, dx$. Evaluating the integral: $I'(b) = \left[ \frac{x^{b+1}}{b+1} \right]_0^1 = \frac{1}{b+1}$. Now,integrate $I'(b)$ with respect to $b$: $I(b) = \int \frac{1}{b+1} \, db = \log(b+1) + C$. When $b = 0$,$I(0) = \int_0^1 \frac{x^0 - 1}{\log x} \, dx = \int_0^1 0 \, dx = 0$. Substituting $b = 0$ into our expression: $I(0) = \log(0+1) + C = 0 \implies C = 0$. Thus,$I(b) = \log(b+1)$.

The value of the integral $\int_0^1 \frac{x^b - 1}{\log x} \, dx$ is

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