-ablog x-xdx- -MP PET 1994-

∫_a^b1/x dx=(log x log x)_a^b ∫_a^b1/x log x dx ⇒ 2I = [(log x)^2]_a^b⇒ I=1/2[(log b)^2 (log a)^2] =1/2[(log b + log a)(log b log a)] = 1/2log(ab)log ( b/a ...

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Byju's Answer

Standard XIII

Mathematics

Second Fundamental Theorem of Calculus

∫ablog x/xdx=...

Question

$\int_{a}^{b} \frac{l o g x}{x} d x =$ [MP PET 1994]

l o g (\frac{l o g b}{l o g a})

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l o g (a b) l o g (\frac{b}{a})

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\frac{1}{2} l o g (a b) l o g (\frac{b}{a})

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\frac{1}{2} l o g (a b) l o g (\frac{a}{b})

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Solution

The correct option is C $\frac{1}{2} l o g (a b) l o g (\frac{b}{a})$
$\int_{a}^{b} \frac{1}{x} d x = (l o g x l o g x)_{a}^{b} - \int_{a}^{b} \frac{1}{x} l o g x d x$
$\Rightarrow 2 I = [(l o g x)^{2}]_{a}^{b} \Rightarrow I = \frac{1}{2} [(l o g b)^{2} - (l o g a)^{2}]$
$= \frac{1}{2} [(l o g b + l o g a) (l o g b - l o g a)] = \frac{1}{2} l o g (a b) l o g (\frac{b}{a})$

Suggest Corrections

∫balog xxdx= [MP PET 1994]

The correct option is C 12log(ab)log(ba)∫ba1x dx=(log x log x)ba−∫ba1xlog x dx ⇒2I=[(log x)2]ba⇒I=12[(log b)2−(log a)2] =12[(log b+log a)(log b−log a)]=12log(ab)log(ba)

$\int_{a}^{b} \frac{l o g x}{x} d x =$ [MP PET 1994]