First slide

Methods of integration

Question

$\int \frac{1 d x}{x (\log x - 2) (\log x - 3)} is equal to$

Moderate

A

$\log (\frac{\log x + 2}{\log x + 3}) + c$
B

$\log (\frac{\log x - \log 3}{\log x - \log 2}) + c$
C

$\log (\frac{\log x - 3}{\log x - 2}) + c$
D

$\log (\frac{\log x + 3}{\log x + 2}) + c$

Solution

$Put \log x = u Then \frac{d x}{x} = d u$
$Hence Given = \int \frac{d x}{x (\log x - 2) (\log x - 3)}$
$= \int \frac{d u}{(u - 2) (u - 3)}$
$Using partial fractions, \frac{1}{(u - 2) (u - 3)} = \frac{A}{u - 2} + \frac{B}{u - 3}$
$1 = A (u - 3) + B (u - 2)$
$Put u = 2 \Rightarrow - 1 = A$
$Put u = 3 \Rightarrow 1 = B$
$\begin{array}{l} = \int \frac{d u}{(u - 2) (u - 3)} = \int [\frac{1}{u - 3} - \frac{1}{u - 2}] d u \\ = \log (u - 3) - \log (u - 2) \\ = \log (\frac{u - 3}{u - 2}) + c \\ = \log (\frac{\log x - 3}{\log x - 2}) + c \end{array}$

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