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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende die Potenzregel an

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

Berechne die Grenzen:

- \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}}

Vereinfache

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

d er i v a t i v e 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