integral from 0 to L of x/(x)

Lösung

\int_{0}^{L} \frac{x}{d + x} d x

- d (ln ∣ d + L ∣ - ln ∣ d ∣) + L

Schritte zur Lösung

\int_{0}^{L} \frac{x}{d + x} d x

\frac{x}{d + x} = - \frac{d}{d + x} + 1

= \int_{0}^{L} - \frac{d}{d + x} + 1 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{L} \frac{d}{d + x} d x + \int_{0}^{L} 1 d x

\int_{0}^{L} \frac{d}{d + x} d x = d (ln ∣ d + L ∣ - ln ∣ d ∣)

\int_{0}^{L} 1 d x = L

= - d (ln ∣ d + L ∣ - ln ∣ d ∣) + L

\frac{\partial}{\partial x} ((2 x + 5 y) e^{y})

\frac{\partial}{\partial x} (x + y - x y)

\int_{0}^{8} (x + 1)^{2} d x

x \to 0 lim (\frac{ln ( 1 - 6 x )}{x})

x \to - 8 lim (\frac{∣ x + 8 ∣}{x + 8})