TITLE: Another beautiful integral (Part 2)
QUESTION [16 upvotes]: One of the ways of calculating the integral in closed form is to think of crafitly using the geometric series, but even so it seems evil enough.
$$\int_0^1\int_0^1\int_0^1\int_0^1\frac{1}{(1+x) (1+y) (1+z)(1+w) (1+ x y z w)} \ dx \ dy \ dz \ dw$$
Maybe you can guide me, bless me with another precious hints, clues. Thanks MSE users! 
Supplementary question: How about the generalization?
$$\int_0^1\int_0^1\cdots\int_0^1\frac{1}{(1+x_1) (1+x_2)\cdots  (1+x_n)(1+ x_1 x_2 \cdots x_n)} \ dx_1 \ dx_2 \cdots \ dx_n$$

REPLY [4 votes]: Here is a physicist's point of view.
As suggested by OP, i will use the simplest, geometric series approach.
Let us now look at the general case.
$$I_n=\int_0^1...\int_0^1\frac{dx_1...dx_n}{(1+x_1)...(1+x_n)(1+x_1...x_n)}$$
Let's use geometric series
$$\frac{1}{1+x_1...x_n}=1+\sum_{\nu=1}^{\infty}(-1)^\nu(x_1...x_n)^\nu$$
Now let's put the last result into $I_n$ and use the following simple result
$$\int_0^1\frac{x^\nu}{1+x}=(-1)^\nu\left [\ln2+\sum_{k=1}^\nu\frac{(-1)^k}{k}  \right ] $$
After some simple calculations(I'll skip them) we reach the end result 
$$I_n=\ln^n2+\sum_{\nu=1}^\infty(-1)^{\nu(n-1)}\left [\ln2+\sum_{k=1}^\nu\frac{(-1)^k}{k}  \right ]^n$$
It is obvious that $I_n$ converges asymptotically to $\ln^n2$. Already at moderate values of $n$, $\ln^n2$ gives a good approximation. 
For example,in case of $n=4$ worked out by David H if we use computed by him value $I_{4}\approx0.223076$,  absolute error, if we use $\ln^42$ instead of $I_4$, is about 0.008