Hint for integration $\int_{-2}^{2} \frac{3x^2}{1+e^x}\mathrm dx$ I found the following problem in JEE-Advanced (2014) question paper:

$$\int_{-2}^{2} \frac{3x^2}{1+e^x}\mathrm dx$$

At present, I don't know how to approach the problem. So, a few good hints would be appreciated.
(It seems there is no closed form for the antiderivative of the function.)
 A: Let $u=-x$.
\begin{align}
\int_{-2}^2\frac{3x^2}{1+e^x}dx&=\int_2^{-2}\frac{-3u^2}{1+e^{-u}}du\\
&=\int_{-2}^2\frac{3u^2}{1+e^{-u}}du\\
&=\int_{-2}^2\frac{3u^2e^u}{1+e^{u}}du\\
&=\int_{-2}^2\frac{3x^2e^x}{1+e^x}dx
\end{align}
So 
$$2\int_{-2}^2\frac{3x^2}{1+e^x}dx=\int_{-2}^2\frac{3x^2}{1+e^x}dx+\int_{-2}^2\frac{3x^2e^x}{1+e^x}dx=3\int_{-2}^2x^2dx$$
A: Although the question has been answered in a similar way, I would like to throw some light on the general strategy for solving definite integrals of this sort. Let us consider $I=\int_a^b f(x) dx$. Substitute $u=a+b-x$. $$I=\int_a^bf(x)dx=-\int_b^af(a+b-x)dx=\int_a^bf(a+b-x)dx \tag1$$Note that this property is useful especially when $f(x)$ is a rational function whose denominator remains the same upon this substitution. Considering the integral the OP has asked. Let $$I=\int_{-2}^2\dfrac{3x^2}{1+e^x}dx=\int_{-2}^{2}\dfrac{3x^2e^x}{1+e^x}dx \implies I=\dfrac{x^3}{3}\mid_0^2=8$$
Note: This answer is written keeping the typology of JEE Definite Integration questions in mind. Definite Integration questions in the JEE usually requires one to make use of formula $(1)$.
A: Here is my own method.
$$\int_{-2}^{2}\frac{3x^2}{1+e^x}\mathrm dx$$
$$=\int_{-2}^{0} \frac{3x^2}{1+e^x}\mathrm dx+ \int_{0}^{2}\frac{3x^2}{1+e^x}\mathrm dx $$
For $x\in[-2,0]$ the fraction $\dfrac{3x^2}{1+e^x}$ becomes $\dfrac{3x^2 e^{-x}}{1+e^{-x}}$
So, $\displaystyle\int_{-2}^{0} \frac{3x^2}{1+e^x}\mathrm dx $ becomes 
$$\int_{0}^{2}\frac{3x^2 e^x}{1+e^x}\mathrm dx$$
Now the expression simplifies,
$$\int_{0}^{2}\frac{3x^2e^x}{1+e^x} \mathrm dx+ \int_{0}^{2}\frac{3x^2}{1+e^x}\mathrm dx=\int_{0}^{2} 3x^2\mathrm dx $$
= 8; which is the answer in my key.
