Evaluate: $$\int_{-2}^{2} \frac{x^2}{1+5^{x}} dx$$
After checking f(-x) for odd/even function and not getting suitable results.
Let $$1+5^x=t$$ $$dx=\frac{dt}{log(5)(t-1)}$$ $$x^2=\frac{(t-1)^2}{(log(5))^2}$$ Limits changed from $[-2,2]$ to $[\frac{26}{25},26]$ $$\int_{\frac{26}{25}}^{26} \frac{(t-1)^2}{[(log(5))^2(t)(log(5)(t-1)]} dt$$ $$\frac{1}{(log(5))^3}\int_{\frac{26}{25}}^{26} \frac{t-1}{t} dt$$ This gave me some abomination that wasn't the answer I required which is 8/3.
I now also know the intended solution as mentioned below:
Let $$I=\int_{-2}^{2} \frac{x^2}{1+5^{x}} dx$$ Using Identity of f(a+b-x) from definite integrals: $$I=\int_{-2}^{2} \frac{(2-2-x)^2}{1+5^{(2-2-x)}} dx$$ $$I=\int_{-2}^{2} \frac{x^2}{1+5^{(-x)}} dx$$ $$I=\int_{-2}^{2} \frac{(5^x)x^2}{1+5^{x}} dx$$ $$2I=\int_{-2}^{2} \frac{x^2}{1+5^{x}}+\frac{(5^x)x^2}{1+5^{x}} dx$$ $$2I=\int_{-2}^{2} x^2 dx$$ $$I=\frac{1}{6}.[2^3-(-2^3)]=\frac{8}{3}$$
So why didn't my first approach work?
Your choice of substitution does not work because if $$t = 1+5^x,$$ then $$x = \log_5 (t-1) = \frac{\color{red}{\log}\, (t-1)}{\log 5}$$ hence $$x^2 = \frac{\color{red}{\log^2 (t-1)}}{\log^2 5}.$$ Consequently, $$\begin{align} \int_{x=-2}^2 \frac{x^2}{1+5^x} \, dx &= \int_{t=26/25}^{26} \frac{1}{t} \cdot \frac{\log^2 (t-1)}{\log^2 5} \cdot \frac{1}{(t-1) \log 5} \, dt \\ &= \frac{1}{\log^3 5} \int_{t=26/25}^{26} \frac{\log^2 (t-1)}{t(t-1)} \, dt, \end{align}$$ and the antiderivative of this integrand does not have an elementary closed form. It can be expressed in terms of the polylogarithm $$\operatorname{Li}_n(z) = \sum_{k=1}^\infty \frac{z^k}{k^n},$$ but this is much too tedious compared to the intended solution.