integral of (29)/(x^3-125)

Lösung

\int \frac{29}{x ^{3} - 125} d x

29 (\frac{1}{75} ln ∣ x - 5 ∣ - \frac{1}{150} ln 4 x^{2} + 20 x + 100 - \frac{1}{25 3} arctan (\frac{2}{5 3} x + \frac{1}{3})) + C

Schritte zur Lösung

\int \frac{29}{x ^{3} - 125} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 29 \cdot \int \frac{1}{x ^{3} - 125} d x

Ermittle den Partialbruch von

\frac{1}{x ^{3} - 125} : \frac{1}{75 ( x - 5 )} + \frac{- x - 10}{75 ( x ^{2} + 5 x + 25 )}

= 29 \cdot \int \frac{1}{75 ( x - 5 )} + \frac{- x - 10}{75 ( x ^{2} + 5 x + 25 )} 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

= 29 (\int \frac{1}{75 ( x - 5 )} d x + \int \frac{- x - 10}{75 ( x ^{2} + 5 x + 25 )} d x)

\int \frac{1}{75 ( x - 5 )} d x = \frac{1}{75} ln ∣ x - 5 ∣

\int \frac{- x - 10}{75 ( x ^{2} + 5 x + 25 )} d x = \frac{1}{150 3} (- 3 ln 4 x^{2} + 20 x + 100 - 6 arctan (\frac{2}{5 3} x + \frac{1}{3}))

= 29 (\frac{1}{75} ln ∣ x - 5 ∣ + \frac{1}{150 3} (- 3 ln 4 x^{2} + 20 x + 100 - 6 arctan (\frac{2}{5 3} x + \frac{1}{3})))

Multipliziere aus

\frac{1}{150 3} (- 3 ln 4 x^{2} + 20 x + 100 - 6 arctan (\frac{2}{5 3} x + \frac{1}{3})) : - \frac{1}{150} ln 4 x^{2} + 20 x + 100 - \frac{1}{25 3} arctan (\frac{2}{5 3} x + \frac{1}{3})

= 29 (\frac{1}{75} ln ∣ x - 5 ∣ - \frac{1}{150} ln 4 x^{2} + 20 x + 100 - \frac{1}{25 3} arctan (\frac{2}{5 3} x + \frac{1}{3}))

Füge eine Konstante zur Lösung hinzu

= 29 (\frac{1}{75} ln ∣ x - 5 ∣ - \frac{1}{150} ln 4 x^{2} + 20 x + 100 - \frac{1}{25 3} arctan (\frac{2}{5 3} x + \frac{1}{3})) + C

\int_{2}^{4} \frac{1}{x x ^{2} - 4} d x

\frac{d}{d x} (\frac{- 2}{x ^{3}})

n = 1 \sum \infty \frac{2 ^{n}}{100}

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

\frac{\partial}{\partial y} (- y sin (x y))