# Integrating $\int^2_{-2}\frac{x^2}{1+5^x}$

$$\int^2_{-2}\frac{x^2}{1+5^x}$$

How do I start to integrate this?

I know the basics and tried substituting $5^x$ by $u$ where by changing the base of logarithm I get $\frac{\ln(u)}{\ln 5}=x$, but I got stuck.

Any hints would suffice preferably in the original question and not after my substitution.

(And also using the basic definite integrals property.)

Now I know only basic integration, that is restricted to high school, so would prefer answer in terms of that level.

• @PabloRotondo I might not know that property ,the only properties I do know are in page 55 of this link [ncert.nic.in/ncerts/l/lemh201.pdf] Mar 14, 2016 at 9:32
• You'll spend a hard time trying to find an antiderivative. Follow @GoodDeeds.
– user65203
Mar 14, 2016 at 9:39
• @MartinSleziak Oh you were linking that question asked today to my question asked 9 months ago. Jan 6, 2017 at 17:52
• Yes, I know. I think that the answers to the question might be useful to people reading your post, so I added a link. (The answers there are a bit more detailed.) Jan 6, 2017 at 18:15

$$\tag1I=\int_{-2}^{2}\frac{x^2}{1+5^x}dx$$ Note that $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ Thus, $$\tag2I=\int_{-2}^{2}\frac{(-2+2-x)^2}{1+5^{-2+2-x}}dx=\int_{-2}^{2}\frac{x^2}{1+5^{-x}}dx=\int_{-2}^{2}\frac{5^xx^2}{1+5^{x}}dx$$

Add $(1)$ and $(2)$.

• So after this I would be left with $x^3$ Right?(for the variable part before appplying upper and lower limit) Mar 14, 2016 at 9:40
• @IshanTaneja No, you will be left with $\frac{x^2}{2}$ in the integral Mar 14, 2016 at 9:40
• I guess that @IshanTaneja means $\frac{x^3}6$, after integration.
– user65203
Mar 14, 2016 at 9:43
• I always get a grin on my face, when I see someone using this kind of "trick" . Even though it is actually so obvious, I rarely see people using it. It's so wonderful :) Mar 14, 2016 at 9:53

Hint:

$$\frac1{1+5^{-x}} + \frac1{1+5^x} = 1$$

• But how do I replace x by -x without any particular property in mind? Mar 14, 2016 at 9:34
• @IshanTaneja: Break the integral up into two pieces, one from $[-2,0]$ and one from $[0,2]$. Sub $x \mapsto -x$ in the former integral. Mar 14, 2016 at 9:36
• Okay I got limit breaking but substituing x by -x will change dx by -dx ,right? Mar 14, 2016 at 9:37
• @IshanTaneja: yes, but that negative will disappear when you reverse the upper and lower limits of that integral. Mar 14, 2016 at 9:38