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

Question

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

https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 — lab bhattacharjee, May 20 '20 at 17:41
https://math.stackexchange.com/questions/540125/how-to-find-this-integral-i-int-pi-pi-fracx-sinx-arctanex1-co/540128#540128 — lab bhattacharjee, May 20 '20 at 17:47

score 10 · Answer 1 · answered May 20 '20 at 17:03

10

Notice that $$I+I=\int_{-2}^{2}\frac{x^{4}}{1+6^{x}}dx+\int_{-2}^{2}\frac{6^{x}x^{4}}{1+6^{x}}dx=\int_{-2}^2x^4dx.$$

answered May 20 '20 at 17:03

Kenta S

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AryanSonwatikar · Answer 2 · 2020-05-21T03:49:39.197

6

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?

edited May 21 '20 at 03:49

answered May 20 '20 at 17:04

AryanSonwatikar

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Typo in denominator? Should be $1 + 6^x$ – Nicholas Roberts May 20 '20 at 17:20
@Nicholas Roberts Yes. Apologies, I'll fix it. – AryanSonwatikar May 21 '20 at 03:49

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

2 Answers2