integral of (x-7)/((x+1)(x^2+2x+5))

Lösung

\int \frac{x - 7}{( x + 1 ) ( x ^{2} + 2 x + 5 )} d x

- 2 ln ∣ x + 1 ∣ + ln x^{2} + 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C

Schritte zur Lösung

\int \frac{x - 7}{( x + 1 ) ( x ^{2} + 2 x + 5 )} d x

Ermittle den Partialbruch von

\frac{x - 7}{( x + 1 ) ( x ^{2} + 2 x + 5 )} : - \frac{2}{x + 1} + \frac{2 x + 3}{x ^{2} + 2 x + 5}

= \int - \frac{2}{x + 1} + \frac{2 x + 3}{x ^{2} + 2 x + 5} 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{2}{x + 1} d x + \int \frac{2 x + 3}{x ^{2} + 2 x + 5} d x

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

\int \frac{2 x + 3}{x ^{2} + 2 x + 5} d x = ln x^{2} + 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x + 1))

= - 2 ln ∣ x + 1 ∣ + ln x^{2} + 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x + 1))

Füge eine Konstante zur Lösung hinzu

= - 2 ln ∣ x + 1 ∣ + ln x^{2} + 2 x + 5 + \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C

\int \frac{cos ( 5 x )}{x} d x

\int cos^{2} (x) d x

\int x arcsin (2 x) d x

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

\int \frac{5 tan ( x ) - 2 cos ^{2} ( x )}{cos ( x )} d x