integral of ((9x+14))/((x-2)(x^2+4))

Lösung

\int \frac{( 9 x + 14 )}{( x - 2 ) ( x ^{2} + 4 )} d x

4 ln ∣ x - 2 ∣ - 2 ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

Schritte zur Lösung

\int \frac{9 x + 14}{( x - 2 ) ( x ^{2} + 4 )} d x

Ermittle den Partialbruch von

\frac{9 x + 14}{( x - 2 ) ( x ^{2} + 4 )} : \frac{4}{x - 2} + \frac{- 4 x + 1}{x ^{2} + 4}

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

\int \frac{4}{x - 2} d x = 4 ln ∣ x - 2 ∣

\int \frac{- 4 x + 1}{x ^{2} + 4} d x = - 2 ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})

= 4 ln ∣ x - 2 ∣ - 2 ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= 4 ln ∣ x - 2 ∣ - 2 ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

\int (2 x + 7)^{3} d x

x \to - 1 lim (\frac{∣ x - 1 ∣}{x - 1})

n = 1 \sum \infty \frac{4}{n} - \frac{4}{n + 1}

\int x^{\frac{1}{12}} d x

\frac{\partial}{\partial x} (\frac{2 x}{x ^{2} - y ^{2}})