how to solve the following definite integral? I am really confused about solving it.
$$\int_{-2}^2 \frac{x^2}{1+5^x} \, dx $$
 A: HINT:
Use $\displaystyle I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$
$\displaystyle I+I=\int_a^bf(x)\ dx+\int_a^bf(a+b-x)\ dx=\int_a^b[f(x)+f(a+b-x)]\ dx$
A: 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
