Integrate $\int \frac{dx}{x \ln x}\, \mathrm{d}x$ using integration by parts

Question

I know that the integral of $\int \frac{1}{x \ln x}\, \mathrm{d}x$ can easily be obtained through substitution for $u=\ln x$ with the result of $\ln \ln x+C$. My question is if this answer (or an equivalent one) can be obtained via integration by parts, and if not why?

I have tried the substitutions $u = \frac{1}{\ln x}$ and $dv = \frac{1}{x}\,dx$ which yeilds $$ \int \frac{1}{x \ln x} \, \mathrm{d}x = 1 + \int \frac{1}{x \ln x} \, \mathrm{d}x$$ which is no good.

Answer 1

So when you do indefinite integration by parts, this equation that you got should actually be

$$ \int \frac{1}{x \ln(x)} \mathrm{d}x = 1 + \int \frac{1}{x \ln(x)} \mathrm{d}x + C$$

or in other words $\int \frac{1}{x \ln(x)}$ represent a family of functions instead of just one function.

Now you might ask, what if you are doing a definite integral?

in the definite integral case, $$ \int^b_a \frac{1}{x \ln(x)} \mathrm{d}x = 1\bigg|^b_a + \int^b_a \frac{1}{x \ln(x)} \mathrm{d}x $$ $$ \int^b_a \frac{1}{x \ln(x)} \mathrm{d}x = (1 - 1) + \int^b_a \frac{1}{x \ln(x)} \mathrm{d}x $$ $$ \int^b_a \frac{1}{x \ln(x)} \mathrm{d}x = \int^b_a \frac{1}{x \ln(x)} \mathrm{d}x $$

so nothing wrong there. To summarize you still want to use $u=ln(x)$ to solve the integral, but integration by parts didn't fail you.

Answer 2

Choose \begin{align} v&=\color{red}{\frac1{\ln x}}\\ du&=\color{blue}{\frac{dx}{x}} \end{align} Thefore we have \begin{align} dv&=\color{green}{-\frac{dx}{x\ln^2 x}}\\ u&=\color{orange}{\ln x} \end{align} Now using the recipe for integrations by part we get\begin{align} \int \frac{dx}{x\ln x}&=\color{red}{\frac1{\ln x}}\times\color{orange}{\ln x}-\int \color{orange}{\ln x}\times \color{green}{(-\frac{dx}{x\ln^2 x})}\\ &=1+\int\frac{dx}{x\ln x} \end{align}

Answer 3

The derivative of ln (x) is 1/x so it's u'/u which is ln (ln(x)) if you want the easiest way to find the integral

Answer 4

It can be done using integration by parts as follows

\begin{align} I=\int \frac{1}{x \ln x}\,{d}x = &\int {2\sqrt{\ln\ln x}}\ d(\sqrt{\ln\ln x})\\ =&\ 2\ln\ln x-(I+C)= \ln \ln x+C’ \end{align} with $C’=-\frac C2$.