TITLE: Evaluation of $\int_0^1 \frac{\log^2(1+x)}{x} \ dx$
QUESTION [18 upvotes]: One of the ways to approach it lies in the area of the dilogarithm, but is it possible to evaluate it
by other means of the real analysis (without using dilogarithm)?
$$\int_0^1 \frac{\log^2(1+x)}{x} \ dx$$  
EDIT: maybe you're aware of some easy way to do that. I'd appreciate it!
Some words on the generalization case (by means of the real analysis again)?
$$F(n)=\int_0^1 \frac{\log^n(1+x)}{x} \ dx, \space n\in \mathbb{N}$$

REPLY [3 votes]: You can find a nice generalization for $\int_0^1\frac{\ln^n(1+x)}{x}dx$ in lemma $2.2$ in this article and I am going to type it here with little more details.
Start with subbing $\frac{1}{1+x}=y$
$$I_n=\int_0^1\frac{\ln^n(1+x)}{x}dx=(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y(1-y)}dy$$
$$=(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y}dy+(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{1-y}dy$$
$$=(-1)^n\left[(-1)^n\frac{\ln^{n+1}(2)}{n+1}\right]+(-1)^n\int_{0}^1\frac{\ln^n(y)}{1-y}dy-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy$$
$$=(-1)^n\left[(-1)^n\frac{\ln^{n+1}(2)}{n+1}\right]+(-1)^n\left[(-1)^n n!\zeta(n+1)\right]-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy$$
$$=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy\tag1$$
By using 
$$(x+y)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$$
or
$$(x-y)^n=(-1)^n(y-x)^n=(-1)^n \sum_{k=0}^n{n\choose k}y^{n-k}(-x)^k=\sum_{k=0}^n{n\choose k}(-y)^{n-k}x^k\tag2$$
we get
$$\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy\overset{2y=x}{=}-\int_0^1\frac{(\ln(x)-\ln(2))^n}{2-x}dx$$
$$\overset{(2)}{=}-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left(\int_0^1\frac{\ln^k(x)}{2-x}dx\right)$$
$$=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left(\sum_{i=1}^\infty\frac1{2^i} \int_0^1 x^{i-1}\ln^k(x)dx\right)$$
$$=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left((-1)^k k!\sum_{i=1}^\infty\frac1{2^i i^{k+1}}\right)$$
$$=-\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}(-1)^k k!\operatorname{Li}_{k+1}\left(\frac12\right)\tag3$$
Plug $(3)$ in $(1)$ we get 

$$I_n=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$
  or
  $$(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y(1-y)}dy=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$