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I am really confused about solving it.

$$\int_{-2}^2 \frac{x^2}{1+5^x} \, dx $$

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closed as off-topic by JMP, Claude Leibovici, user230715, Hans Lundmark, gebruiker Mar 22 '16 at 10:31

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    $\begingroup$ Your first step (from $\int_{-2}^2$ to $2\int_0^2$) is invalid, since the integrand is not symmetric ($x^2$ is symmetric, but $1+5^x$ is not). $\endgroup$ – Arthur Mar 22 '16 at 8:02
  • $\begingroup$ @Arthur ok, sorry my bad, let me make a quick edit. $\endgroup$ – Gautam Chandna Mar 22 '16 at 8:06
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    $\begingroup$ You don't have to delete your attempt, because it is sufficiently showing your effort. $\endgroup$ – choco_addicted Mar 22 '16 at 8:09
  • $\begingroup$ I doubt there is an easy analytical solution. For a numerical one, though, try it out on wolfram alpha. $\endgroup$ – Vim Mar 22 '16 at 8:12
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    $\begingroup$ Possible duplicate of Integrating $\int^2_{-2}\frac{x^2}{1+5^x}$ $\endgroup$ – Hans Lundmark Mar 22 '16 at 9:59
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HINT:

Use $\displaystyle I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

$\displaystyle I+I=\int_a^bf(x)\ dx+\int_a^bf(a+b-x)\ dx=\int_a^b[f(x)+f(a+b-x)]\ dx$

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There is a solution in special functions, achieved via use of substitutions.

$\int_{-2}^2 \frac{x^2}{1+5^x} \, dx$

Let $t=5^x$

$x= \log_5(t)$

$dx = \frac{dt}{\log_5(t) \ln(5)}$

$t(-2)=\frac{1}{25}; t(2)=25$

$\int_{-2}^2 \frac{x^2}{1+5^x} \, dx = $$\int_{\frac{1}{25}}^{25} \frac{(\log_5(t))^2}{1+t} \, \frac{dt}{\log_5(t) \ln(5)}$

$\int_{\frac{1}{25}}^{25} \frac{\log_5(t)}{1+t} \, \frac{dt}{ \ln(5)}$

$\frac{1}{ (\ln(5))^2} \int_{\frac{1}{25}}^{25} \frac{\ln(t)}{1+t} \, dt$

This last integral is special. It is closely related to the polylogarithm, and in particular, the dilogarithm.

By using the properties of these functions and the regular logarithm, I am sure you could reason out exactly what WolframAlpha tells us this is equal to:

$\frac{1}{ (\ln(5))^2} [Li_2(-t)+\log(t)\log(t+1)]^{25}_{\frac{1}{25}}$

$\frac{1}{ (\ln(5))^2} [[Li_2(-25)+\log(25)\log(26)]-[Li_2(-\frac{1}{25})+\log(\frac{1}{25})\log(\frac{26}{25})]]$ ≈1.493

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  • $\begingroup$ Sorry, let me elaborate..one sec. $\endgroup$ – KR136 Mar 22 '16 at 8:20
  • $\begingroup$ Thats awesome! Learned some new stuff from your answer. Thanks alot! $\endgroup$ – Gautam Chandna Mar 22 '16 at 10:38
  • $\begingroup$ @GautamChandna No problem! Substitutions always surprise me- they so often are more useful than you think they'd be. $\endgroup$ – KR136 Mar 22 '16 at 13:15
  • $\begingroup$ Your solution gave me a new way to look at such problems! Thanks for it. $\endgroup$ – Gautam Chandna Mar 22 '16 at 18:49

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