# Integral of $\frac{1}{x\ln(x+1)}$

I'm trying to get my head around calculating $$\int\frac{1}{x\ln(x+1)}dx.$$ I can't seem to get anywhere. I tried parts and substitutions, but that $(x+1)$ is always in the way.

Any suggestions?

EDIT:
Thanks to all those who have pointed out that there is no simple expression for the integral. Does anyone know a closed-form approximation? I'm not interested in a definite integral, but rather in an approximation to $$\int_1^x\frac{1}{t\ln(t+1)}dt.$$

• This integral is actually not computable if you check Wolfram Mathematica. Mar 15, 2014 at 16:35
• Mathematica 8 returns this unevaluated. Mar 15, 2014 at 16:36
• Fairly often an antiderivative cannot be expressed in terms of the standard elementary functions. Some evidence that this might be the case in particular instance can be obtained by asking Wolfram Alpha, or Mathematica, or Maple to do it. Mar 15, 2014 at 16:36
• Are you looking for a numerical approximation over a given interval? Mar 15, 2014 at 16:36
• I suppose $\int_1^x \frac{1}{t \ln(t+1)} \; dt$ is not a satisfactory answer here. Mar 15, 2014 at 16:39

Choose a number $N$ that is relatively large, like 10 or 20 or what have you, and then you might try $$\int_1^x{{dt\over t \ln\!{(t+1)}}} \approx \int_1^N{dt\over t\ln\!(t+1)} + \int_N^x{dt\over t\ln t}\qquad \qquad (x \geq N)$$ The first integral on the right-hand side can be evaluated numerically given the choice of $N$, and the second integral is $\ln\!{(\ln{x}/\ln{N})}$.

So for example, the choice $N = 20$ leads to the following approximation for $x \geq 20$: $$\int_1^x{{dt\over t \ln\!{(t+1)}}} \approx 0.9415 + \ln\!{(\ln{x}})\,.$$

• Nice! This might work!
– Mau
Mar 15, 2014 at 21:09
• If you want to use $N = 100$, I made a table of values for the integral for $x = 1$ to $x = 100$. dropbox.com/s/bqbnrxysq100oz2/DataTable.csv Mar 15, 2014 at 23:15

Let $u=\ln(x+1)$ ,

Then $x=e^u-1$

$dx=e^u~du$

$\therefore\int\dfrac{1}{x\ln(x+1)}dx$

$=\int\dfrac{e^u}{u(e^u-1)}du$

$=\int\dfrac{1}{u(1-e^{-u})}du$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nB_nu^{n-2}}{n!}du$ (with the formula in http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function)

$=\int\left(\dfrac{1}{u^2}-\dfrac{1}{2u}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_nu^{n-2}}{n!}\right)du$

$=-\dfrac{1}{u}-\dfrac{\ln u}{2}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_nu^{n-1}}{n!(n-1)}+C$

$=-\dfrac{1}{\ln(x+1)}-\dfrac{\ln\ln(x+1)}{2}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_n(\ln(x+1))^{n-1}}{n!(n-1)}+C$

$\therefore\int_1^x\dfrac{1}{t\ln(t+1)}dx$

$=\left[-\dfrac{1}{\ln(t+1)}-\dfrac{\ln\ln(t+1)}{2}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_n(\ln(t+1))^{n-1}}{n!(n-1)}\right]_1^x$

$=\dfrac{1}{\ln2}-\dfrac{1}{\ln(x+1)}+\dfrac{\ln\ln2}{2}-\dfrac{\ln\ln(x+1)}{2}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_n((\ln(x+1))^{n-1}-(\ln2)^{n-1})}{n!(n-1)}$