$\int \frac{d x}{x (l o g x - 2) (l o g x - 3)} = f (x) + c t h e n f (x) =$

$\frac{1}{x} l o g \frac{| l o g x - 3 |}{| l o g x - 2 |}$

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$l o g \frac{| l o g x - 3 |}{| l o g x - 2 |}$

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$2 l o g \frac{| l o g x - 3 |}{| l o g x - 2 |}$

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$l o g | (l o g x - 3) (l o g x - 2) |$

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Solution

The correct option is B
$l o g \frac{| l o g x - 3 |}{| l o g x - 2 |}$

$P u t l o g x = t$
$\Rightarrow I = \int \frac{1}{(t - 2) (t - 3)} d t$
$I = l o g ∣ ∣ \frac{l o g x - 3}{l o g x - 2} ∣ ∣ + c$

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Similar questions

\int \frac{d x}{x (log x - 2) (log x - 3)} =

\int sec (log x) (1 + tan (log x)) d x = f (x) + C

f (x) = x sec (log x)

\int {x f^{'} (x) + f (x)} d x = x f (x) + C

f (x) = x log x

$\int \frac{d x}{x (l o g x - 2) (l o g x - 3)} = f (x) + c t h e n f (x) =$

f (x) = {tan}^{- 1} x - (1 / 2) log x

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