First slide

Methods of integration

Question

If $\int [\log (\log x) + \frac{1}{(\log x)^{2}}] d x$
$= x [f (x) - g (x)] + C$ , then

Moderate

A

$f (x) = \log (\log x); g (x) = \frac{1}{\log x}$
B

$f (x) = \log x; g (x) = \frac{1}{\log x}$
C

$f (x) = \frac{1}{\log x}; g (x) = \log (\log x)$
D

$f (x) = \frac{1}{x \log x}; g (x) = \frac{1}{\log x}$

Solution

$\int \log (\log x) d x$
$\begin{array}{l} = x \log (\log x) - \int x \frac{1}{\log x} \cdot \frac{1}{x} d x \\ = x \log (\log x) - \int \frac{d x}{\log x} \\ = x \log (\log x) - x \frac{1}{\log x} + \int x \frac{(- 1)}{(\log x)^{2}} \frac{1}{x} d x \\ \Rightarrow \int [\log (\log x) + \frac{1}{(\log x)^{2}}] d x = x [f (x) - g (x)] + C \end{array}$
where $f (x) = \log (\log x)$
and $g (x) = \frac{1}{\log x}$

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