Question

${\int }_a^b\frac{logx}{x}dx=$ [MP PET 1994]

• A. $log\left(\frac{logb}{loga}\right)$
• B. $log\left(ab\right)log\left(\frac{b}{a}\right)$
• C. $\frac{1}{2}log\left(ab\right)log\left(\frac{b}{a}\right)$
• D. $\frac{1}{2}log\left(ab\right)log\left(\frac{a}{b}\right)$

Answer 1

The correct option is C $\frac{1}{2}log\left(ab\right)log\left(\frac{b}{a}\right)$

${\int }_a^b\frac{1}{x}dx=(logxlogx{)}_a^b-{\int }_a^b\frac{1}{x}logxdx$
$\Rightarrow 2I=\left[\right(logx{)}^2{]}_a^b\Rightarrow I=\frac{1}{2}\left[\right(logb{)}^2-\left(loga{)}^2\right]$
$=\frac{1}{2}\left[\right(logb+loga\left)\right(logb-loga\left)\right]=\frac{1}{2}log\left(ab\right)log\left(\frac{b}{a}\right)$