Question:
Evaluate the following integrals
$\int \frac{1}{x \log x} d x$
Solution:
Assume $\log x=t$
$\mathrm{d}(\log x)=\mathrm{dt}$
$\frac{1}{x} d x=d t$
Put $\mathrm{t}$ and $\mathrm{dt}$ in given equation we get
$\Rightarrow \int \frac{d t}{t}$
$=\ln |\mathrm{t}|+\mathrm{c}$
But $t=\log x$
$=\ln |\log x|+c$
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