Question

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Answer 1

$\displaystyle V=\left\{(x,y,z): 0\leq x\leq 1\,; \,0\leq y\leq 1-x\,; \,0\leq z\leq 1-x-y\right\}\\ \\ H=\iiint_{V}\frac{dx\,dy\,dz}{(x+y+z+1)^3}\\ \\ \\ H=\int\limits_{0}^{1}dx \int\limits_{0}^{1-x}dy\int\limits_{0}^{1-x-y}\frac{1}{(x+y+z+1)^3}dz\\ \\ \\ H=\int\limits_{0}^{1}dx \int\limits_{0}^{1-x}\left.\frac{-1/2}{(x+y+z+1)^2}\right|_{z=0}^{z=1-x-y}dy$

$\displaystyle H=-\frac{1}{2}\int\limits_{0}^{1}dx \int\limits_{0}^{1-x}\frac{1}{4}-\frac{1}{(x+y+1)^2}dy\\ \\ \\ H=-\frac{1}{2}\int\limits_{0}^{1}\left.\left(\frac{y}{4}+\frac{1}{x+y+1}\right)\right|_{0}^{1-x}dx \\ \\ \\ H=-\frac{1}{2}\int\limits_{0}^{1}\left(\frac{1-x}{4}+\frac{1}{2}-\frac{1}{x+1}\right)dx \\ \\ \\ H=-\frac{1}{2}\int\limits_{0}^{1}\left(\frac{3}{4}-\frac{x}{4}-\frac{1}{x+1}\right)dx$

$\displaystyle H=-\frac{1}{2}\left.\left(\frac{3}{4}x-\frac{x^2}{8}-\ln|x+1|\right)\right|_0^1\\ \\ \\ H=-\frac{1}{2}\left(\frac{3}{4}-\frac{1}{8}-\ln2\right)\\ \\ \\ \boxed{H=\ln \sqrt{2}-\frac{5}{16}}$