Why does the whole integral converge but not part of it? (Dilogs) $\newcommand{\Li}{\operatorname{Li}}$Consider the integral:
$$\int_0^1 \frac{(-\Li_2(x) - \Li_3(x) - x^2/8 + 3x - x\log(1-x) + \log(1-x))}{x^2} \, dx$$
This integral converges to $\sim 0.01$
But when taken separately, 
$$ \int_0^1 \frac{-\Li_2(x) - \Li_3(x)}{x^2} \, dx$$
this integral does not converge.
What is going on here? I dont understand? 
 A: Hint:
After subtracting out the quadratic term in the numerator, we're left with three terms in the numerator whose power series expansions each begin with a $-x$ we need to somehow cancel, plus $3x$. The key is then distributing these three $x$'s to pair up with one these other terms and then we may split the sum:
$$\begin{align}
I
&=\int_{0}^{1}\frac{3x-\frac{x^2}{8}+\left(1-x\right)\ln{\left(1-x\right)}-\operatorname{Li}_{2}{\left(x\right)}-\operatorname{Li}_{3}{\left(x\right)}}{x^2}\,\mathrm{d}x\\
&=-\frac18+\int_{0}^{1}\frac{3x+\left(1-x\right)\ln{\left(1-x\right)}-\operatorname{Li}_{2}{\left(x\right)}-\operatorname{Li}_{3}{\left(x\right)}}{x^2}\,\mathrm{d}x\\
&=-\frac18+\int_{0}^{1}\frac{x+\left(1-x\right)\ln{\left(1-x\right)}+x-\operatorname{Li}_{2}{\left(x\right)}+x-\operatorname{Li}_{3}{\left(x\right)}}{x^2}\,\mathrm{d}x\\
&=-\frac18+\int_{0}^{1}\frac{x+\left(1-x\right)\ln{\left(1-x\right)}}{x^2}\,\mathrm{d}x+\int_{0}^{1}\frac{x-\operatorname{Li}_{2}{\left(x\right)}}{x^2}\,\mathrm{d}x+\int_{0}^{1}\frac{x-\operatorname{Li}_{3}{\left(x\right)}}{x^2}\,\mathrm{d}x.\\
\end{align}$$
