<s>[INST] What's the method of integrating $\frac{1}{1-\ln(x)}$? How to do $\int\frac{1}{1-\ln(x)}dx$ manually? I mean I got its answer by running it through Mathematica, but is there anyway we can do it by hand?
 [/INST] $\frac{1}{1 - \ln x}$ has no elementary antiderivative. If you are so inclined you could express $1 - \ln x$ as a series.
$$\ln (1+x) = x - x^2/2 + x^3/3 - ...$$
$$\to \ln (1+x) = (-1+1+x) - (-1+1+x)^2/2 + (-1+1+x)^3/3 - ...$$
$$\to \ln (x) = (-1+x) - (-1+x)^2/2 + (-1+x)^3/3 - ...$$
$$\to -\ln (x) = -(-1+x) + (-1+x)^2/2 - (-1+x)^3/3 - ...$$
$$\to 1-\ln (x) = 1-(-1+x) + (-1+x)^2/2 - (-1+x)^3/3 - ...$$
$$\to 1-\ln (x) = 1+(1-x) + (-1+x)^2/2 - (-1+x)^3/3 - ...$$
$$\to 1-\ln (x) = 1+(x-1) + (x-1)^2/2 - (x-1)^3/3 - ...$$

Read more:
http://en.wikipedia.org/wiki/List_of_integrals_of_logarithmic_functions
http://en.wikipedia.org/wiki/Mercator_series

Alternatively:
Let $x=e^{1-z}, dx = -e^{1-z} dz$ where we can write:
$$e^{1-z} = \sum_{n=0}^{\infty} \frac{(1-z)^n}{n!}$$
$$\int\frac{dx}{1-\ln x}=\int\frac{-e^{1-z} }{z}dz = \int\frac{-1}{z}\sum_{n=0}^{\infty} \frac{(1-z)^n}{n!}dz = \sum_{n=0}^{\infty} \int\frac{-1}{z}\frac{(1-z)^n}{n!}dz$$
Since we can write
$$(1-z)^n = \sum_{k=0}^{n} \binom n k (-z)^k$$
we have
$$= \sum_{n=0}^{\infty} \int\frac{-1}{z}\frac{(1-z)^n}{n!}dz = -\sum_{n=0}^{\infty} \sum_{k=0}^{n} [\frac{1}{k!(n-k)!}(\frac{z^k}{k} + C)]$$
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