# Solving the definite integral in closed form [duplicate]

The following question was asked during the final exam of AISSCE. Evaluate $\int_{-2}^2 \frac{x^2}{1 + 5^x}dx$. I think it must be easy and leave it to wide audience to check...cheers!

## marked as duplicate by GoodDeeds, Martin Sleziak, gebruiker, user1551, Alex ProvostMar 16 '16 at 22:19

• Hint: substitute $x \to -x$ and consider the sum of the new and the original integral. – StackTD Mar 16 '16 at 11:31
Note that $\int_{-2}^2\frac{x^2}{1+5^x}dx=\int_{0}^2\frac{x^2}{1+5^x}dx+\int_{-2}^0\frac{x^2}{1+5^x}dx$. Taking $u=-x$ in the second integral, we have
$\begin{eqnarray}\int_{-2}^2\frac{x^2}{1+5^x}\,\mathrm{d}x&=&\int_{0}^2\frac{x^2}{1+5^x}\,\mathrm{d}x+\int_{-2}^0\frac{x^2}{1+5^x}\,\mathrm{d}x\\&=&\int_{0}^2\frac{x^2}{1+5^x}\,\mathrm{d}x+\int_{0}^2\frac{u^25^u}{1+5^u}\,\mathrm{d}u\\&=&\int_0^2\frac{x^2(1+5^x)}{1+5^x}\,\mathrm{d}x\\&=&\int_0^2x^2\,\mathrm{d}x\\&=&\frac{8}{3}\end{eqnarray}$