integral of 1/((t+4)(t-1))

Lösung

\int \frac{1}{( t + 4 ) ( t - 1 )} d t

- \frac{1}{5} ln ∣ t + 4 ∣ + \frac{1}{5} ln ∣ t - 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{( t + 4 ) ( t - 1 )} d t

Ermittle den Partialbruch von

\frac{1}{( t + 4 ) ( t - 1 )} : - \frac{1}{5 ( t + 4 )} + \frac{1}{5 ( t - 1 )}

= \int - \frac{1}{5 ( t + 4 )} + \frac{1}{5 ( t - 1 )} d t

Wende die Summenregel an:

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

= - \int \frac{1}{5 ( t + 4 )} d t + \int \frac{1}{5 ( t - 1 )} d t

\int \frac{1}{5 ( t + 4 )} d t = \frac{1}{5} ln ∣ t + 4 ∣

\int \frac{1}{5 ( t - 1 )} d t = \frac{1}{5} ln ∣ t - 1 ∣

= - \frac{1}{5} ln ∣ t + 4 ∣ + \frac{1}{5} ln ∣ t - 1 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{5} ln ∣ t + 4 ∣ + \frac{1}{5} ln ∣ t - 1 ∣ + C

(x y - y^{2}) d x - x^{2} d y = 0

\frac{\partial}{\partial x} (- 5 y^{2} sin (x y^{2}))

\frac{\partial}{\partial y} (20 (y - x^{2}))

x \to 1 lim (\frac{x ^{2} + 1}{x + 5})

(x + y)^{2} d x + (2 x y + x^{2} - 6) d y = 0, y (1) = 1