Question

The value of integral $\int_{0}^{1}\frac{x^{b} - 1}{\log x}dx$ is

• A. $\log b$
• B. $2\log(b + 1)$
• C. $3\log b$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

Let $I(b) = \int_{0}^{1}\frac{x^{b} - 1}{\log x}dx \Rightarrow I'(b) = \int_{0}^{1}{\frac{x^{b}\log x}{\log x}dx}$

(If $I(\alpha) = \int_{0}^{b}{f(x,\alpha)dx}$, then $I'(\alpha) = \int_{0}^{b}{f'(x,\alpha)dx}$, where $f'(x,\alpha)$ is derivative of $f(x,\alpha)$ w.r.t. $\alpha$ keeping x constant)

$I^{'}(b) = \int_{0}^{1}{x^{b}dx = \frac{1}{b + 1}}$

⇒ $I(b) = \int_{}^{}{\frac{db}{b + 1} + c = \log(b + 1) + c}$

If $b = 0$, then $I(b) = 0$, so $c = 0$⇒$I(b) = \log(b + 1)$.