How do you find the triple integral? I need some help finding the triple integral of: 1/xyz. All three integration limits are from 1 to e^2. Help is appreciated! Thank you in advance!
 A: $$
\int_1^{e^2} \int_1^{e^2} \left(\int_1^{e^2} \frac 1 x\cdot\underbrace{\frac 1 y \cdot \frac 1 z} \, dx\right)\,dy\,dz
$$
The part over the $\underbrace{\text{underbrace}}$ does not change as $x$ goes from $1$ to $e^2$; and it is a factor, i.e. the thing being integrated is the part over the $\underbrace{\text{underbrace}}$ multiplied by something.  Hence that part can be pulled out:
$$
\int_1^{e^2} \int_1^{e^2} \left(\underbrace{\frac 1 y \cdot \frac 1 z}\int_1^{e^2} \frac 1 x \, dx\right)\,dy\,dz
$$
Doing this innermorst integral, you get a number, $2$.
Then iterate as needed.
A: First you need to know, that when integrating or derivating multivariables, the variable that you ARE NOT operating counts as a constant, so:
$$ \int_1^{e^2}\int_1^{e^2}\int_1^{e^2}\frac{1}{x}\frac{1}{y}\frac{1}{z} \,dx\,dy\,dz = \int_1^{e^2}\int_1^{e^2} \frac{1}{y}\frac{1}{z} \left(\int_1^{e^2}\frac{1}{x}\,dx \right )\,dy\,dz = \int_1^{e^2}\int_1^{e^2} \frac{1}{y}\frac{1}{z} \Big[ \ln(x) \Big]_1^{e^2} \, dy\, dz $$
$$ \Big[ \ln(x) \Big]_1^{e^2} = \ln(1) -\ln(e^2) = 0 - 2\ln(e) = 2 $$
$$ = 2\int_1^{e^2} \frac{1}{z} \left( \int_1^{e^2} \frac{1}{y}\,dy \right )\, dz = (\cdots) $$
I hope I have helped, 
Saclyr.
