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


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*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?

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.
