Problem in computing of integral by substitution. I want to compute an integral like 
$\int_0^{+\infty} \ln(1+x)e^{-x}\,\mathrm dx$. 
Then denote $\mu = 1-e^{-x}$, so $x=-\ln(1-\mu)$. Substitute this into the integral, we get

$$\int_0^1 \ln(1-\ln(1-\mu))(1-\mu)\,\mathrm d(-\ln(1-\mu)) = \int_0^1 \ln(1-\ln(1-\mu))\,\mathrm d\mu$$

However, $\ln(1-\ln(1-\mu))$ goes to infinity if $\mu$ goes to 1, which makes the latter integral uncomputable. But the original integral has no such problem. Is there any error in my substitution?
You reduce the integral over the infinite interval to the one over the finite interval. You shall not be surprised that the function under the integral may become unbounded. The point is that you may think of the value of the integral as an area under the graph of the function. Now, what you change a variable, you make a correspondent change of the coordinate system where they area is defined. That shall give you an intuition why the function may (but doesn't have to) become unbounded. However, it still stays integrable. 

Your operations with $\mu$ seem fine to me. To support this, I computed both integral in Mathematica (see the image below), they're both well-defined and have the same value.

For some reason, Wolfram Alpha computes only an approximate value of the second integral.
we have 
$$\int_{0}^{\infty}e^{\mu x}\ln{(\beta+x)}dx=\dfrac{1}{\mu}[\ln{\beta}-e^{\mu\beta}Ei{(-\beta\mu)}]$$