integral of 1/(P-P^2)

Lösung

\int \frac{1}{P - P ^{2}} d P

ln ∣ P ∣ - ln ∣ P - 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{P - P ^{2}} d P

Ermittle den Partialbruch von

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

= \int \frac{1}{P} - \frac{1}{P - 1} d P

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}{P} d P - \int \frac{1}{P - 1} d P

\int \frac{1}{P} d P = ln ∣ P ∣

\int \frac{1}{P - 1} d P = ln ∣ P - 1 ∣

= ln ∣ P ∣ - ln ∣ P - 1 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ P ∣ - ln ∣ P - 1 ∣ + C

\int \frac{x}{x + 4} d x

t an g e n t f (x) = \frac{2 x}{x ^{2} - 7}, (3, 3)

(x^{2} + 1) y^{^{'}} + 9 x (y - 1) = 0

\int (cos (x))^{4} d x

d er i v a t i v e f (x) = \frac{x - 3}{x + 4}