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

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First Fundamental Theorem of Calculus

∫ab log x/xdx...

Question

$\int_{a}^{b} \frac{l o g x}{x} d x =$

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 $\frac{1}{2} l o g (a b) l o g (\frac{b}{a})$
$L e t I = \int_{a}^{b} \frac{1}{x} l o g x d x = (l o g x l o g x)_{a}^{b} - \int_{a}^{b} \frac{1}{x} l o g x d x$
$\Rightarrow 2 I = [(l o g x)^{2}]_{a}^{b} \Rightarrow I = \frac{1}{2} [(l o g b)^{2} - (l o g a)^{2}]$
$\Rightarrow \frac{1}{2} [(l o g b + l o g a) (l o g b - l o g a)] = \frac{1}{2} l o g (a b) l o g (\frac{b}{a})$

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∫balog xxdx=

The correct option is C 12log(ab)log(ba)Let I=∫ba1xlog x 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)

$\int_{a}^{b} \frac{l o g x}{x} d x =$