Solve a definite integral involving exponentiation Solve the following integral:
$$\int_0^1 \frac{x}{3^x+3^{1-x}-3}dx$$
I can't think of any useful substitution or any rewriting so that integration by parts could be applied.
Thank you in advance!
 A: HINT:
$$I = \int_0^1 f(x) dx = \int_0^1 f(1-x)dx$$
$$\implies I = \frac{1}{2}\int_0^1(f(x) + f(1-x)) dx = \frac{1}{2}\int_0^1\frac{dx}{3^x + 3^{1-x} -3}$$
An appropriate substitution can be made to solve this integral.
A: Hint. By the change of variable $u=1-x$ one gets
$$
\int_0^1 \frac{x}{3^x+3^{1-x}-3}\:dx=\int_0^1 \frac{1-u}{3^u+3^{1-u}-3}\:du.
$$
Can you finish it?
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\int_{0}^{1}{x \over 3^{x} + 3^{1 - x} - 3}\,\dd x =
\int_{0}^{1}{3^{-x}\,x \over 3 \times 3^{-2x} -3 \times 3^{-x} + 1}\,\dd x
\\[5mm] \stackrel{t\ =\ 3^{-x}}{=}\,\,\, &
{1 \over 3}\int_{1}^{1/3}{t\bracks{-\ln\pars{t}/\ln\pars{3}} \over
t^{2} - t + 1/3}\,\bracks{-\,{1 \over \ln\pars{3}}\,{1 \over t}}\,\dd t =
-\,{1 \over 3\ln^{2}\pars{3}}\int_{1/3}^{1}
{\ln\pars{t} \over \pars{t - r}\pars{t - \bar{r}}}\,\dd t
\end{align}
where
$\ds{r \equiv {1 \over 2} + {\root{3} \over 6}\,\ic =
{\root{3} \over 3}\expo{\pi\ic/6}}$.

Then,
\begin{align}
&\int_{0}^{1}{x \over 3^{x} + 3^{1 - x} - 3}\,\dd x =
-\,{1 \over 3\ln^{2}\pars{3}}\int_{1/3}^{1}\ln\pars{t}
\pars{{1 \over t - r} - {1 \over t - \bar{r}}}{1 \over r - \bar{r}}\,\dd t
\\[5mm] = &\
{1 \over 3\ln^{2}\pars{3}\pars{\root{3}/6}}\,
\Im\int_{1/3}^{1}{\ln\pars{t} \over r - t}\,\dd t =
{2\root{3} \over 3\ln^{2}\pars{3}}\,
\Im\int_{1/\pars{3r}}^{1/r}{\ln\pars{rt} \over 1 - t}
\,\dd t
\\[5mm] = &\
{2\root{3} \over 3\ln^{2}\pars{3}}\,
\Im\bracks{%
\ln\pars{1 - {1 \over 3r}}\ln\pars{r\,{1 \over 3r}} +
\int_{1/\pars{3r}}^{1/r}{\ln\pars{1 - t} \over t}\,\dd t}
\\[5mm] = &\
{2\root{3} \over 3\ln^{2}\pars{3}}\,
\Im\bracks{%
-\ln\pars{1 - {1 \over 3r}}\ln\pars{3} -
\mrm{Li}_{2}\pars{1 \over r} + \mrm{Li}_{2}\pars{1 \over 3r}}
\\[5mm] = &\
\bbox[20px,border:1px dotted navy]{\ds{{2\root{3} \over 3\ln^{2}\pars{3}}\,
\bracks{-\,{\ln\pars{3} \over 6}\,\pi -
\Im\mrm{Li}_{2}\pars{3 - \root{3}\ic \over 2} + 
\Im\mrm{Li}_{2}\pars{3 - \root{3}\ic \over 6}}}} \approx 0.8255
\end{align}
