integral of (65x+7)/((8x+1)(x-1))

Lösung

\int \frac{65 x + 7}{( 8 x + 1 ) ( x - 1 )} d x

\frac{1}{8} ln ∣ 8 x + 1 ∣ + 8 ln ∣ x - 1 ∣ + C

Schritte zur Lösung

\int \frac{65 x + 7}{( 8 x + 1 ) ( x - 1 )} d x

Ermittle den Partialbruch von

\frac{65 x + 7}{( 8 x + 1 ) ( x - 1 )} : \frac{1}{8 x + 1} + \frac{8}{x - 1}

= \int \frac{1}{8 x + 1} + \frac{8}{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}{8 x + 1} d x + \int \frac{8}{x - 1} d x

\int \frac{1}{8 x + 1} d x = \frac{1}{8} ln ∣ 8 x + 1 ∣

\int \frac{8}{x - 1} d x = 8 ln ∣ x - 1 ∣

= \frac{1}{8} ln ∣ 8 x + 1 ∣ + 8 ln ∣ x - 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} ln ∣ 8 x + 1 ∣ + 8 ln ∣ x - 1 ∣ + C

\int (sin (x) \cdot cos (x)) d x

y^{^{''}} - 2 y^{^{'}} + y = 3 e^{2 x}

t an g e n t y = \frac{4}{x}, (- 5, - \frac{4}{5})

7 x y \frac{d y}{d x} = 1

s l o p e (- 5, - 7), (4, - 1)