integral of 1/((x-2)(x+4))

Lösung

\int \frac{1}{( x - 2 ) ( x + 4 )} d x

\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣ + C

Schritte zur Lösung

\int \frac{1}{( x - 2 ) ( x + 4 )} d x

Ermittle den Partialbruch von

\frac{1}{( x - 2 ) ( x + 4 )} : \frac{1}{6 ( x - 2 )} - \frac{1}{6 ( x + 4 )}

= \int \frac{1}{6 ( x - 2 )} - \frac{1}{6 ( x + 4 )} 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}{6 ( x - 2 )} d x - \int \frac{1}{6 ( x + 4 )} d x

\int \frac{1}{6 ( x - 2 )} d x = \frac{1}{6} ln ∣ x - 2 ∣

\int \frac{1}{6 ( x + 4 )} d x = \frac{1}{6} ln ∣ x + 4 ∣

= \frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 4 ∣ + C

\int - \frac{9000}{( 3 x + 50 ) ^{2}} d x

\int \frac{1}{y + 5} d y

\frac{\partial}{\partial x} (yz + x z + x y)

\frac{\partial}{\partial x} (8 \cdot y^{22})

y^{^{''}} - y^{^{'}} = (3 x^{2} + 6 x + 1) e^{x}