What is the integral from 0 to y+z of 1/((x+y+z)^3) ?

The integral from 0 to y+z of 1/((x+y+z)^3) is 3/(2(2y+2z)^2)

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integral from 0 to y+z of 1/((x+y+z)^3)

\int_{0}^{y + z} \frac{1}{( x + y + z ) ^{3}} d x

\frac{3}{2 ( 2 y + 2 z ) ^{2}}

Solution steps

\int_{0}^{y + z} \frac{1}{( x + y + z ) ^{3}} d x

Apply u-substitution

: \int_{y + z}^{2 y + 2 z} \frac{1}{u ^{3}} d u

= \int_{y + z}^{2 y + 2 z} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{y + z}^{2 y + 2 z}

= [- \frac{1}{2 u ^{2}}]_{y + z}^{2 y + 2 z}

Compute the boundaries:

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

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

Simplify

= \frac{3}{2 ( 2 y + 2 z ) ^{2}}

e^{x} (x - 1)

y^{^{'}} + sin (t) y = sin (t), y (0) = 6

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

\frac{d}{d x} ((e^{3 x}) (cos (2 x)))

16 y^{''} + 8 y^{'} + y = 0

What is the integral from 0 to y+z of 1/((x+y+z)^3) ?
The integral from 0 to y+z of 1/((x+y+z)^3) is 3/(2(2y+2z)^2)