How to evaluate $\int_{1}^{2}\frac{dx}{1+x+\ln x}$? Can you help me find the value of the integral 

$$\int_{1}^{2}\frac{dx}{1+x+\ln x}$$

Thank you
 A: According to Maple, there is no closed form either for the antiderivative or the definite integral.  In this "pure transcendental" case,  Maple's implementation of the Risch algorithm is complete, so there is no elementary antiderivative (this could also be done using the Rothstein-Trager theorem, see e.g. http://www.math.ubc.ca/~israel/m210/lesson18.pdf ).  That doesn't mean there can't be an elementary formula for some definite integrals, but since there doesn't appear to be anything special about $2$ in this context, it's not very likely.  The numerical value .35700808127536096106 isn't recognized by Maple's identify or Plouffe's Inverter.
A: $\int_1^2\dfrac{dx}{1+x+\ln x}$
$=\int_0^{\ln2}\dfrac{d(e^x)}{1+e^x+x}$
$=\int_0^{\ln2}\dfrac{e^x}{e^x+x+1}dx$
$=\int_0^{\ln2}\dfrac{1}{1+(x+1)e^{-x}}dx$
$=\int_0^{\ln2}\left(1+\sum\limits_{n=1}^\infty(-1)^n(x+1)^ne^{-nx}\right)dx$
$=\left[x-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(n-1)!(x+1)^ke^{-nx}}{n^{n-k}k!}\right]_0^{\ln2}$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$=\ln2-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(n-1)!(\ln2+1)^k}{2^nn^{n-k}k!}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(n-1)!}{n^{n-k}k!}$
A: I would make the substitution $x=a+1$, that one can actually get a power series sans logarithms:
$$\int_{1}^{2} \frac{dx}{1+x+\log(x)} = \int_{0}^{1} \frac{da}{2+a+\log(1+a)}$$
One may be able to find sequences on oeis that this matches up with, I was not.
(aside: entertainingly the first five terms for the numerators of the power series of $f(a)=\frac{1}{2+a+\log(1+a)}$ match up with oeis:A048607 (Numerators of coefficients in function $a(x)$ such that $a(a(x))$ = $\ln(1+x)$), but the fifth does not, -497  vs -749 for $f(a)$)
