integral of 1/((x-3)(x-1))

Lösung

\int \frac{1}{( x - 3 ) ( x - 1 )} d x

\frac{1}{2} ln ∣ x - 3 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{( x - 3 ) ( x - 1 )} d x

Ermittle den Partialbruch von

\frac{1}{( x - 3 ) ( x - 1 )} : \frac{1}{2 ( x - 3 )} - \frac{1}{2 ( x - 1 )}

= \int \frac{1}{2 ( x - 3 )} - \frac{1}{2 ( 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 \frac{1}{2 ( x - 3 )} d x - \int \frac{1}{2 ( x - 1 )} d x

\int \frac{1}{2 ( x - 3 )} d x = \frac{1}{2} ln ∣ x - 3 ∣

\int \frac{1}{2 ( x - 1 )} d x = \frac{1}{2} ln ∣ x - 1 ∣

= \frac{1}{2} ln ∣ x - 3 ∣ - \frac{1}{2} ln ∣ x - 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln ∣ x - 3 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + C

d er i v a t i v e \frac{6}{5} - \frac{6}{5} e^{- 20 t}

l o n g d i v i s i o n \frac{x ^{2} + 1}{x ^{2} - 4}

\frac{d y}{d x} = 4 - y

\int \frac{x ^{3}}{9} d x

d er i v a t i v e f (x) = \frac{7 x}{3 + x ^{2}}