Triple integral of e^(x+y+z)

I = ∫₀^{ln 2} ∫₀^x ∫₀^{x + ln y} e^(x+y+z) dz dy dx

(The upper limit of z is x + ln y. I evaluate the integral exactly as printed.)

z-integration

∫₀^{x+ln y} e^(x+y) e^z dz = e^(x+y) [ e^(x + ln y) − 1 ]
                           = e^(x+y) ( y eˣ − 1 )
                           = y e^(2x+y) − e^(x+y)

y-integration (from 0 to x)

∫₀ˣ y e^(2x+y) dy = e^(2x) [ (y − 1)eʸ ]₀ˣ = e^(2x) [ (x − 1)eˣ + 1 ]
∫₀ˣ e^(x+y) dy    = eˣ (eˣ − 1) = e^(2x) − eˣ

Subtract:

(x − 1)e^(3x) + e^(2x) − e^(2x) + eˣ = (x − 1)e^(3x) + eˣ

x-integration (from 0 to ln 2)

Using ∫ (x − 1)e^(3x) dx = (x − 1)e^(3x)/3 − e^(3x)/9, with e^(3 ln 2) = 8:

[(x − 1)e^(3x)/3 − e^(3x)/9] at ln 2 = 8(ln 2 − 1)/3 − 8/9
                             at 0    = −1/3 − 1/9 = −4/9
difference = 8 ln 2/3 − 8/3 − 8/9 + 4/9 = 8 ln 2/3 − 28/9

∫₀^{ln 2} eˣ dx = 1

I = 8 ln 2/3 − 28/9 + 1 = 8 ln 2/3 − 19/9

Answer: I = (8/3) ln 2 − 19/9 ≈ −0.263.

(The value is negative because for small y the upper limit x + ln y is negative, so the z-integral runs backwards there.)