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Evaluate $$\int_{-\pi/6}^{\pi/6}\frac{\tan^{-1}x}{\mathrm{e}^x+1} \,\mathrm{d} x $$

My try: I used integration by parts using $\tan^{-1}x$ as the first function and $\displaystyle \frac{1}{\mathrm{e}^x+1}$ as second function but again i faced with a difficult integral with Integrand:

$$\frac{\log_{e}(e^x+1)}{1+x^2}$$

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    $\begingroup$ Are you sure it should not be $\arctan x^2$, $\arctan |x|$ or anything along those lines? If $f(x) = f(-x)$ then $$ \int_{-a}^a \frac{f(x)}{e^x+1}\,\mathrm{d}x = \int_{0}^a f(x) \,\mathrm{d}x $$ Because $$ \int_{-a}^a f(x)\mathrm{d}x = \int_0^a f(x) + f(-x) \mathrm{d}x $$ $\endgroup$ – N3buchadnezzar May 4 '14 at 8:11
  • $\begingroup$ @N3buchadnezzar, Yes i am sure its not what you expected... $\endgroup$ – Ekaveera Kumar Sharma May 4 '14 at 8:33
  • $\begingroup$ Where did this integral come from? Do you have a good reason to expect it to have a closed form in terms of well-known functions and constants? $\endgroup$ – user111187 May 4 '14 at 9:10
  • $\begingroup$ The output doesn't seem to be too rational...this integral seems to be coming out of nowhere, and just random stuff...wolframalpha says, the integral is $0.0224439...$ $\endgroup$ – Hawk May 4 '14 at 10:33
  • $\begingroup$ What is its answer sir,i am curious to know?@EkaveeraKumarSharma $\endgroup$ – Brahmagupta Feb 6 '16 at 3:49
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For now, we may use series; since

$$\frac{1}{1+e^x} = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{e^{nx}}},$$ valid for $|e^x|<1$, and

$$\tan^{-1}x = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{2n + 1}}{x^{2n + 1}}}, $$ valid for $|x| <1$, then the series

$$\sum\limits_{n = 0}^\infty {\sum\limits_{k = 0}^n {\frac{{{{\left( { - 1} \right)}^{k + n}}}}{{2k + 1}}{x^{2k + 1}}} } {e^{nx}}$$, valid for $|xe^x|<r<1$

and converges uniformly there. So that \begin{align} \int {\frac{{ta{n^{ - 1}}x}}{{1 + {e^x}}}dx} &= \int {\left( {\sum\limits_{n = 0}^\infty {\sum\limits_{k = 0}^n {\frac{{{{\left( { - 1} \right)}^{k + n}}}}{{2k + 1}}{x^{2k + 1}}} } {e^{nx}}} \right)dx} \\ &= \sum\limits_{n = 0}^\infty {\sum\limits_{k = 0}^n {\frac{{{{\left( { - 1} \right)}^{k + n}}}}{{2k + 1}}} } \left( {\int {{x^{2k + 1}}{e^{nx}}dx} } \right) \end{align} and of course the integral ${\int {{x^{2k + 1}}{e^{nx}}dx} }$ follows by parts.

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