First slide

Evaluation of definite integrals

Question

The value of the integral $\int_{0}^{1} \frac{x^{c} - 1}{\log x} d x$ , c > 0 is

Moderate

A

log c
B

$2 \log (c + 1)$
C

3 log c
D

$\log (c + 1)$

Solution

Note that the integral is not an elementary function.
Let $I (c) = \int_{0}^{1} \frac{x^{c} - 1}{\log x} d x$ so
$I^{'} (c) = \int_{0}^{1} \frac{x^{c} \log x d x}{\log x} = \int_{0}^{1} x^{c} d x = \frac{1}{c + 1} .$
Hence $I (c) = \int \frac{1}{c + 1} d c + C = \log | c + 1 | + C .$
But $I (0) = 0$ , hence C=0 so $I (c) = \log (c + 1)$

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