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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = 1 + y + z + x

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

Add a constant to the solution

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

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