Is there a closed form expression for integral $\int_{0}^a \frac{1}{x} \ln(x+1) dx$? Is there a closed form result for integral
$$
\int_{0}^a \frac{1}{x} \ln(x+1) dx
$$
where $a$ is a positive real value.
 A: I don't know if this counts as a "closed form", but if $a\in(0,1)$ we have:
$$ \int_{0}^{a}\frac{\log(1+x)}{x}\,dx = \int_{0}^{a}\sum_{n\geq 1}\frac{(-1)^{n-1} x^{n-1}}{n}\,dx = \sum_{n\geq 1}\frac{(-1)^{n-1} a^n}{n^2}. \tag{1}$$
that can be written as $-\text{Li}_2(-a)=\text{Li}_2(a)-\frac{1}{2}\text{Li}_2(a^2).$
By taking the limit as $\alpha\to 1^-$ we also have $\int_{0}^{1}\frac{\log(1+x)}{x}\,dx=\frac{\pi^2}{12}=\eta(2)=\frac{\zeta(2)}{2}.$ 
If $a>1$, we may use the dilogarithm reflection formulas to get:
$$ \int_{0}^{a}\frac{\log(1+x)}{x}\,dx = \frac{\pi^2}{6}+\frac{\log^2(a)}{2}+\sum_{n\geq 1}\frac{(-1)^{n}}{n^2 a^n}.\tag{2} $$
Equivalently,
$$\begin{eqnarray*} \int_{0}^{a}\frac{\log(1+x)}{x}\,dx &=& \frac{\pi^2}{12}+\int_{1}^{a}\frac{\log(1+x)}{x}\,dx\\&=&\frac{\pi^2}{12}+\int_{\frac{1}{a}}^{1}\frac{\log(x+1)-\log(x)}{x}\,dx\\&=&\frac{\pi^2}{6}+\frac{\log^2(a)}{2}-\int_{0}^{\frac{1}{a}}\frac{\log(1+x)}{x}\,dx \end{eqnarray*}$$
proves the above reflection formula $(2)$ through $(1)$.
