Exercise 1792Calculus › Multivariable calculus › Multiple integralsL1Added 3 Nov 2026

Triple integral over a cube

#cube#triple integral#iterated

Statement

Compute $\displaystyle I=\iiint_{[0,1]^3}\frac{dx\,dy\,dz}{(x+y+z+1)^3}$ .

Hint

Integrate in $x$ first: $\int_0^1\frac{dx}{(x+c)^3}=\frac{1}{2}\left(\frac{1}{c^2}-\frac{1}{(c+1)^2}\right)$ .

Show solutionHide solution

Inner in $x$ . $\frac12\left(\frac{1}{(y+z+1)^2}-\frac{1}{(y+z+2)^2}\right)$ .

In $y$ . $\frac12\left(\frac{1}{z+1}-\frac{2}{z+2}+\frac{1}{z+3}\right)$ .

In $z$ . $I=\frac12\left(\ln2-2\ln\frac32+\ln\frac43\right)=\frac{5\ln2-3\ln3}{2}$ .

Conclusion. $I=\frac{5\ln2-3\ln3}{2}$ .