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

Lösung

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

- ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

= \int - \frac{1}{x + 2} + \frac{1}{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}{x + 2} d x + \int \frac{1}{x + 1} d x

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

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

= - ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣

Füge eine Konstante zur Lösung hinzu

= - ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣ + C

\int \frac{x}{4 x ^{2} + 1} d x

\int \frac{sinh ( x )}{e ^{x} + 1} d x

\int \frac{1}{( x - 3 ) x ^{2} - 6 x + 6} d x

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

\int \frac{1}{cos ^{4} ( x ) sin ^{2} ( x )} d x