∫_-2^2(x^2)/(1+5^x) dx is equal to:

The correct answer is Option (3) → 8/3

Let I be the integral. By the property \( \int_{-a}^{a}f(x)\,dx=\int_{-a}^{a}f(-x)\,dx \),

    \( \displaystyle I=\int_{-2}^{2}\frac{(-x)^2}{1+5^{-x}}\,dx \)

    \( \displaystyle =\int_{-2}^{2}\frac{x^2\cdot 5^x}{5^x+1}\,dx \)

Adding this to the original form of I:

    \( \displaystyle 2I=\int_{-2}^{2}\frac{x^2(1+5^x)}{1+5^x}\,dx \)

    \( \displaystyle =\int_{-2}^{2}x^2\,dx \)

    \( \displaystyle =\left[\frac{x^3}{3}\right]_{-2}^{2} \)

    \( \displaystyle =\frac{8}{3}+\frac{8}{3}=\frac{16}{3} \)

    \( \displaystyle I=\frac{8}{3} \)

Tip: For any a > 0 and even g, \( \int_{-k}^{k}\frac{g(x)}{1+a^x}\,dx=\int_{0}^{k}g(x)\,dx \); the base a does not matter.