Cheeky double integral

$\begin{aligned} I & = \int_1^e \int_{-2}^2 \frac {x^2 \ln y}{y+e^x y}dx\ dy \\ & = \int_1^e \frac {\ln y}y {\color{#3D99F6} \int_{-2}^2 \frac {x^2}{1+e^x}dx}\ dy & \small \color{#3D99F6} \text{See proof: } \int_{-a}^a \frac {f(x)}{1+c^{g(x)}} dx = \int_0^a f(x)\ dx \\ & = \int_1^e \frac {\ln y}y {\color{#3D99F6} \int_0^2 x^2\ dx}\ dy & \small \color{#3D99F6} \text{where }f(x) \text{ is even and }g(x) \text{ odd.} \\ & = \color{#3D99F6} \frac 83 \color{#D61F06} \int_1^e \frac {\ln y}y \ dy & \small \color{#D61F06} \text{By integration by parts} \\ & = \frac 83 \color{#D61F06} \left(\ln^2y \ \bigg|_1^e - \int_1^e \frac {\ln y}y \ dy \right) & \small \color{#D61F06} \text{Note that } \int_1^e \frac {\ln y}y \ dy = \ln^2y \ \bigg|_1^e - \int_1^e \frac {\ln y}y \ dy \\ & = \frac 83 \times \color{#D61F06} \frac 12 \bigg[\ln^2y \bigg]_1^e \\ & = \frac 83 \times {\color{#D61F06} \frac 12} = \frac 43 \end{aligned}$

Therefore, $a+b = 4+3 = \boxed 7$ .

---

Proof for $\small \displaystyle \int_{-a}^a \frac {f(x)}{1+c^{g(x)}} dx = \int_0^a f(x)\ dx$ for even $f(x)$ and odd $g(x)$ .

The given double integral after some rearrangement is:

$I = \int_{1}^{e} \int_{-2}^{2} \left(\frac{x^2}{1+e^x}\right)\left(\frac{\ln(y)}{y}\right)dxdy$

Now, since the limits of integration are constant, the integral can be rewritten as:

$I = \left(\int_{1}^{e}\frac{\ln(y)}{y}dy\right)\left(\int_{-2}^{2}\frac{x^2}{1+e^x}dx\right) = I_1I_2$

$I_1 = \int_{1}^{e}\frac{\ln(y)}{y}dy$

Taking $\ln(y) = z$ transforms the integral to:

$I_1 = \int_{0}^{1}z dz = \frac{1}{2}$

Now:

$I_2 = \int_{-2}^{2}\frac{x^2}{1+e^x}dx = \int_{-2}^{2}\frac{x^2}{1+e^{-x}}dx$

Therefore:

$2I_2 = \int_{-2}^{2}x^2 dx \implies I_2 = \frac{8}{3}$

Finally:

$\boxed{I = I_1I_2 = \frac{4}{3}}$

Note:

If $a$ , $b$ , $c$ and $d$ are constants, then:

$\boxed{J = \int_{a}^{b} \int_{c}^{d} f(x)g(y)dxdy = \left(\int_{a}^{b}g(y)dy\right)\left(\int_{c}^{d} f(x)dx\right)}$