5
$\begingroup$

$\int_{-2}^{2} \frac{x^2}{1+5^x}dx$

I am stuck at the first step and have tried replacing $5^x$ with $e^{\ln(5^x)}$ but nothing simplifies out in the end.

Any hints how I should proceed?

  • 0
    Perhaps try $x=log_5 u$, then integrating by parts. Not sure if that will work though. – 2017-01-05
  • 1
    See also [Integrating $\int^2_{-2}\frac{x^2}{1+5^x}$](http://math.stackexchange.com/q/1696811) and other posts [linked there](http://math.stackexchange.com/questions/linked/1696811). Found [using Approach0](https://approach0.xyz/search/?q=%24%5Cint_%7B-2%7D%5E%7B2%7D%20%5Cfrac%7Bx%5E2%7D%7B1%2B5%5Ex%7Ddx%24&p=1). – 2017-01-06

2 Answers 2

14

$$I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx$$

Let $y=-x$, then:

$$I=\int_{-2}^{2} \frac{y^2}{1+5^{-y}}dy=\int_{-2}^{2} \frac{y^25^y}{1+5^y}dy=\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx$$

So:

$$I+I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx+\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx=\int_{-2}^2x^2\,dx=2\int_0^2x^2\,dx=2\left(\frac{2^3}{3}\right)=\frac{16}{3}$$

So, the integral is equal to $\dfrac{8}{3}$.

Edit: A clarification on the changing of limits under $y=-x$:

$$\int_{x=-2}^{x=2}f(x)\,dx=\int_{-y=-2}^{-y=2}f(-y)(-\,dy)=-\int_{y=2}^{y=-2}f(-y)\,dy=\int_{y=-2}^{y=2}f(-y)\,dy$$

$$=\int_{-2}^{2}f(-y)\,dy$$

  • 0
    With $y = -x$, $dy = -dx$, wouldn't this change the integral function as negative sign will be multiplied? – 2017-01-05
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    @lulu Can you explain a bit more on how this works? Thanks! – 2017-01-05
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    @Gauz Actually, what I said was incorrect (and I will delete my comment in a minute). The point is that the sign changes from the differential, as you point out, but the limits are also inverted and those two sign corrections cancel. Thus, what's posted here is correct. – 2017-01-05
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    Why not -ve sign. And no limit changing when putting y=-x? – 2017-01-06
7

$$ \int_{-2}^{2}\frac{x^2}{1+5^x}\,dx = \int_{-2}^{0}\frac{t^2}{1+5^t}\,dt + \int_{0}^{2}\frac{x^2}{1+5^x}\,dx $$ Substitute $t=-x$ in the first integral to get $$ \int_{0}^{2}\frac{x^25^x}{1+5^x}\,dx $$

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    why are we allowed to change the variable for only 1st part of the integral? I've never seen this actually. – 2017-01-05
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    @Gauz They're two distinct integrals, why shouldn't you be allowed to do it? – 2017-01-06