How do I evaluate $\int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z} \frac{1}{(x+y+z)^3}\text{d}x\text{d}y\text{d}z$?

Question

$$\int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z} \frac{1}{(x+y+z)^3}\text{d}x\text{d}y\text{d}z$$

This is the answer WolframAlpha gives:

substituting it with polar coordinates makes it worse..

Answer 1

You just need to evaluate the integrals from inside. When you are integrating with respect to $x$, the other variables ($y$ and $z$) are constants: $$\int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z} \frac{1}{(x+y+z)^3}\text{d}x\text{d}y\text{d}z=$$$$\int_{1}^{2} \int_{0}^{z}\left.-\frac{1}{2(x+y+z)^2}\right|_{x=0}^{x=y+z}\text{d}y\text{d}z=$$$$-\frac{1}{2}\int_{1}^{2} \int_{0}^{z}\frac{1}{(2y+2z)^2}-\frac{1}{(y+z)^2}\text{d}y\text{d}z=$$$$-\frac{1}{2}\int_{1}^{2} \int_{0}^{z}\frac{1}{2^2}\frac{1}{(y+z)^2}-\frac{1}{(y+z)^2}\text{d}y\text{d}z=$$$$\frac{3}{8}\int_{1}^{2} \int_{0}^{z}\frac{1}{(y+z)^2}\text{d}y\text{d}z=$$$$\frac{3}{8}\int_{1}^{2}\left.-\frac{1}{y+z}\right|_{y=0}^{y=z}\text{d}z=$$$$-\frac{3}{8}\int_{1}^{2}\frac{1}{2z}-\frac{1}{z}\text{d}z=$$$$\frac{3}{16}\int_{1}^{2}\frac{1}{z}\text{d}z=$$$$\frac{3}{16}\log(2)$$