Definite integral $\int_{-2}^{2} \frac{x^2}{1+5^x}\,\mathrm{d}x$

Question

The question is to find the value of the definite integral: $$I=\int_{-2}^{2} \frac{x^2}{1+5^x}\,\mathrm{d}x.$$

This question appeared in this year's CBSE board exam.

Attempts:

Replace $x\to-x$ and we get $$I=\int_{-2}^{2} 5^x \frac{x^2}{1+5^x}\,\mathrm{d}x.$$ Now what?? Probably integration by parts? But I get stuck there too.
Substitute $5^x=\tan^2\theta$, and then I get (after calculation): $$I=\left(\frac2{\log5}\right)^3\int_{\tan^{-1}(1/5)}^{\tan^{-1}(5)}\frac{(\log\tan\theta)^2}{\tan\theta}\,\mathrm{d}x.$$ But I still dont get the answer. It has become too complicated.
Wolfram|Alpha gives me the value $$I=\frac83,$$ But I don't have a pro account so I can't see the steps.

So how do I solve it?

@DhanushKrishna OMG! Why didn't I think of it before! Thanks! — Kartik, Mar 25 '16 at 16:44
This is at least the fourth time this has been asked here in the last two weeks... — Hans Lundmark, Mar 25 '16 at 19:37

score 8 · Accepted Answer · answered Mar 25 '16 at 16:46

8

After substituting $x$ as $-x$ , add the two equations.

answered Mar 25 '16 at 16:46

Dhanush Krishna

score 0 · Answer 2 · answered Mar 25 '16 at 16:44

0

After point 1) integrate by parts using

$$f'(x) = \frac{5^x}{5^x + 1} ~~~~~~~~~~~ \to ~~~~~ f(x) = \frac{\ln(5^x + 1)}{\ln(5)}$$

$$g(x) = x^2 ~~~~~~~~~~~ \to ~~~~~ g'(x) = 2x$$

Then again by parts once.

answered Mar 25 '16 at 16:44

Enrico M.

2 Answers2