What is the integral of 1/((x+y+z+1)^3) ?

The integral of 1/((x+y+z+1)^3) is -1/(2(1+x+y+z)^2)+C

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integral of 1/((x+y+z+1)^3)

\int \frac{1}{( x + y + z + 1 ) ^{3}} d z

- \frac{1}{2 ( 1 + x + y + z ) ^{2}} + C

Solution steps

\int \frac{1}{( x + y + z + 1 ) ^{3}} d z

Apply u-substitution

= \int \frac{1}{u ^{3}} d u

Apply the Power Rule

= - \frac{1}{2 u ^{2}}

Substitute back

u = 1 + x + y + z

= - \frac{1}{2 ( 1 + x + y + z ) ^{2}}

Add a constant to the solution

= - \frac{1}{2 ( 1 + x + y + z ) ^{2}} + C

x \to 0 lim (\frac{2. 7 ^{x} - 1}{x})

\frac{d}{d x} (3 x^{2} + 4 x - 5)

t a y l or e^{x} - 1

x \to 1 lim ((x - 1)^{2})

\frac{\partial}{\partial x} (x^{3} + 2 x y - y^{2})

What is the integral of 1/((x+y+z+1)^3) ?
The integral of 1/((x+y+z+1)^3) is -1/(2(1+x+y+z)^2)+C