_a^b x x ,dx = - Vidyadip

MCQ

$\int_a^b {\frac{{\log x}}{x}\,dx = } $

A
$\log \left( {\frac{{\log b}}{{\log a}}} \right)$
B
$\log (a\,b)\log \,\left( {\frac{b}{a}} \right)$
✓
$\frac{1}{2}\log (a\,b)\log \,\left( {\frac{b}{a}} \right)$
D
$\frac{1}{2}\log (a\,b)\log \,\left( {\frac{a}{b}} \right)$

✓

Answer

Correct option: C.

$\frac{1}{2}\log (a\,b)\log \,\left( {\frac{b}{a}} \right)$

c
(c) Let $I = \int_a^b {\frac{1}{x}\log x\,dx = (\log x\log x)_a^b} $$ - \int_a^b {\frac{1}{x}\log x\,dx} $

$ \Rightarrow 2I = [{(\log x)^2}]_a^b$

$\Rightarrow I = \frac{1}{2}[{(\log b)^2} - {(\log a)^2}]$

$ = \frac{1}{2}[(\log b + \log a)(\log b - \log a)]$

$=\frac{1}{2}\log (ab)\log \left( {\frac{b}{a}} \right)$.

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