Definite Integral involving reciprocals of logs

Question

Integrate $$\int_2^{4e} \frac{1}{x \ln(x+1)}\,dx $$

I have tried partial fractions, u substitution and parts but i cant get the final answer out. my main problem is dealing with the $x$ and $x+1$ simultaneously.

Integrating by partial fractions, i am left trying to integrate $1/\ln(u)$ which i can no longer remember how to do.

My question is; what is the easiest way to calculate the integral and also $1/\ln(u)$. Also please provide information on why you did what you did (why specific substitutions were chosen etc.) and any general strategies for integrals like these.

Answer 1

There is no easy way to express $$\int \frac{dx}{\ln x},$$ you need special functions for that. You can try to integrate it as a series (within the radius of convergence).

Answer 2

This last integral is so particular that it is a special function; it is the definition of the logarithmic integral function $$\text{li}(x)=\int_0^x \frac {dt}{\log(t)}$$ Concerning $$\int_2^{4e} \frac{1}{x \ln(x+1)}\,dx$$ I really do not see how to compute it in a closed form (I did not find any way to get the antiderivative). I am afraid that, probably, numerical integration could be required.

The problem would have been simple with $$I=\int \frac{1}{y \ln(y)}\,dy=\log (\log (y))$$ Using the last one, we could have showed that $$\log \left(\frac{\log (1+4 e)}{\log (3)}\right)<\int_2^{4e} \frac{1}{x \ln(x+1)}\,dx<\log \left(\frac{1+\log (4)}{\log (2)}\right)$$ and, surprizingly, the numerical value of the integral $(\approx 1.02081)$ is very close to the average of the two boundind expressions $(\approx 0.811901$ and $\approx 1.23625)$.