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Question Number 35018 by ajfour last updated on 14/May/18

$$\int \int \int \frac{dxdydz}{{(x+y+z+1)}^3}boundedbythe$$$$coordinateplanesandtheplane$$$$x+y+z=1.$$

Answered by ajfour last updated on 14/May/18

$${\int }_0^1\left[{\int }_0^{1-x}\left[{\int }_0^{1-x-y}\frac{dz}{{(x+y+z+1)}^3}\right]dy\right]dx$$$$={\int }_0^1\left[{\int }_0^{1-x}{(-\frac{1}{2{(x+y+z+1)}^2})}_0^{1-x-y}\right]dy]dx$$$$=\frac{1}{2}{\int }_0^1\left[{\int }_0^{1-x}(\frac{1}{{(x+y+1)}^2}-\frac{1}{4})dy\right]dx$$$$=\frac{1}{2}{\int }_0^1{[\frac{-1}{x+y+1}-\frac{y}{4}]}_0^{1-x}dx$$$$=\frac{1}{2}{\int }_0^1(\frac{1}{x+1}-\frac{1}{2}-\frac{1-x}{4})dx$$$$=\frac{1}{2}{[\ln (x+1)-\frac{3x}{4}+\frac{x^2}{8}]}_0^1$$$$=\frac{\ln 2}{2}-\frac{5}{16}.$$

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