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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

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

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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?

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    $\begingroup$ Typo in denominator? Should be $1 + 6^x$ $\endgroup$ – Nicholas Roberts May 20 '20 at 17:20
  • $\begingroup$ @Nicholas Roberts Yes. Apologies, I'll fix it. $\endgroup$ – AryanSonwatikar May 21 '20 at 3:49

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