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Evaluate$\int^1_0 \int^{1-x}_0 \int^{1-x-y}_0 \frac{1}{(x + y + z +1)^3} dz \ dy \ dx$
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Solution:

$\begin{aligned} \text { Let } 1 &=\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \frac{1}{(x+y+z+1)^{3}} d x \ d y \ d z \\ &=\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \frac{1}{(x+y+z+1)^{3}} d z \ d y \ d x \end{aligned}$

$\begin{aligned} &=\int_{0}^{1} \int_{0}^{1-x}\left[\frac{1}{-2(x+y+z+1)^{-2}}\right]_0^{1-x-y} d y d x \\ &=-\int_{0}^{1} \int_{0}^{1-x} \frac{1}{2}\left[\frac{1}{(x+y+1-x-y+1)^{2}}-\frac{1}{(x+y+1)^{2}}\right] d y d x \end{aligned}$

$\begin{aligned} &=-\int_{0}^{1} \frac{1}{2}\left[\frac{1}{4} y+\frac{1}{(x+y+1)^{1}}\right]_{0}^{1-x} d x \\ &=\int_{0}^{1} \frac{1}{2}\left\{\left[\frac{1}{4}(1-x)-\frac{1}{2}\right]+\left[\frac{1}{x+1}\right]\right\} d x \\ &=\frac{1}{2}\left[\frac{1}{4}\left(\frac{(1-x)^{2}}{8}\right)-\frac{x}{2}+\log (x+1)\right]_{0}^{1} \\ \therefore I &=\frac{1}{2}\left[\log 2-\frac{5}{8}\right] \end{aligned}$

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