Question

Evaluate :

${\int }_0^1\frac{1}{2x-3}dx$

• A. $\frac{1}{3}\log 2$
• B. $-\frac{1}{2}\log 3$
• C. $\log 3$
• D. None of these

Answer 1

The correct option is B $-\frac{1}{2}\log 3$

${\int }_0^1\frac{1}{2x-3}dx$

$=\frac{1}{2}\left[\log \right(2x-3){]}_0^1$

$=\frac{1}{2}[\log |-1|-\log |-3|]$

$=\frac{1}{2}(\log 1-\log 3)$

$=\frac{1}{2}(0-\log 3)$

$=-\frac{1}{2}\log 3$