Computing space intergral

Question

Problem

Compute following:

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

Attempt to solve

Now i could probably solving this by opening $(z+y+x)^3$ parenthesis and then solving 3 separate integrals involving variables $x,y,z$. Opening these parenthesis wouldn't obviously look nice and it would be a lot of work for sure. I think better way would be to substitute $u=(z+y+x)$ and then we would need to adjust the integration limits by working chainrule backwards ?

$$ \frac{\delta u}{\delta x}=1, \quad \delta u = \delta x $$ $$ \frac{\delta u}{\delta y}=1, \quad \delta u = \delta y $$ $$ \frac{\delta u}{\delta z}=1, \quad \delta u = \delta z $$

$$ \int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z}\frac{1}{(z+y+x)^3}\delta x\delta y\delta z $$ $$ \int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z}\frac{1}{u^3}\delta u^3$$ $$\int_{1}^{2} \int_{0}^{z} -\frac{1}{2(y+x)^2}\delta u^2$$

Now it is visible at this point that this cant possibly work.I don't think i have very good understanding on how i am suppose to accomplish this what i am trying to do.

Answer 1

Hint: $$ \int\frac{1}{(x+a)^3}dx=-\frac{1}{2(x+a)^2}+c $$

So:

$$ \int_0^{y+z}\frac{1}{(z+y+x)^3} dx=-\frac{1}{2(z+y+z+y)^2}+\frac{1}{2(z+y)^2}=\frac{1}{4(z+y)^2} $$ and the other integrations are similar.

Answer 2

You just need to repeatedly use the fact that $\int \frac{1}{(v+c)^n} dv = \frac{1}{1-n} \frac{1}{(v+c)^{n-1}} + C$