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

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

Apply the Power Rule

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

= - \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

\frac{x}{( 2 + x )}

sin (x) sec^{2} (x)

x^{3} - 2 x^{2} - x + 1

\frac{4}{3 x - 2}

\frac{1}{( 1 + cos ^{2} ( x ) )}

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