What is the antiderivative of $\frac{1}{x ln x}$ $1/(xlnx)$ ?

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Answer 1 · 2015-09-26T11:51:52

$ln (ln x)$ $ln(lnx)$

antiderivative of $\frac{1}{x ln x}$ $1/(xlnx)$ could be written as $\int (\frac{1}{x ln x} d x)$ $int(1/(xlnx)dx)$

Or as $\int \frac{1}{ln x} \cdot \frac{1}{x} d x$ $int1/(lnx)*1/xdx$

Recall that : the derivative of $ln x$ $lnx$ is $\frac{1}{x}$ $1/x$
This means that : $\frac{d (ln x)}{d x} = \frac{1}{x}$ $(d(lnx))/(dx)=1/x$

Implying : $d (ln x) = \frac{1}{x} (d x)$ $d(lnx)=1/x(dx)$

Hence, $\int \frac{1}{ln x} \cdot \frac{1}{x} d x = \int \frac{1}{ln x} \cdot d (ln x)$ $int1/lnx*1/xdx=int1/lnx*d(lnx)$

we are left with the antiderivative of $\frac{1}{ln x}$ $1/lnx$ with respect to $ln x$ $lnx$

This case is simlar to finding the antiderivative of say, $\frac{1}{u}$ $1/u$ wrt $u$ $u$
The answer would be $ln u$ $lnu$

Similarly, the antiderivative of $\frac{1}{ln x}$ $1/lnx$ with respect to $ln x$ $lnx$ is $ln (ln x)$ $color(blue)(ln(lnx))$

What is the antiderivative of 1xlnx 1/(xlnx)?