# Evaluate $\displaystyle\int\limits_{-2}^{2}\frac{x^{4}}{1+6^{x}}dx$

Problem:

Compute $$I=\displaystyle\int_{-2}^{2}\frac{x^{4}}{1+6^{x}}dx$$

Wolfram Alpha gave me :

$$I=\frac{32}{5}$$

I used $$y=-x$$ and then integral became:

$$I=\displaystyle\int\limits_{-2}^{2}\frac{6^{x}x^{4}}{1+6^{x}}dx$$ But I don't know how to complete, I don't have no ideas.

I am waiting for your solution.

Thanks

Notice that $$I+I=\int_{-2}^{2}\frac{x^{4}}{1+6^{x}}dx+\int_{-2}^{2}\frac{6^{x}x^{4}}{1+6^{x}}dx=\int_{-2}^2x^4dx.$$
You are going in the right direction. Add the $$2$$ different expressions of $$I$$ and you get $$2I=\int_{-2}^2 \frac{(1+6^x)x^4}{1+6^x} dx=\int_{-2}^2 x^4 dx$$ Can you finish?
• Typo in denominator? Should be $1 + 6^x$ – Nicholas Roberts May 20 '20 at 17:20