Solve a definite integral involving exponentiation

Question

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!

Answer 1

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.

Answer 2

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?

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{r \equiv {1 \over 2} + {\root{3} \over 6}\,\ic = {\root{3} \over 3}\expo{\pi\ic/6}}$.

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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}