integral of (\sqrt[3]{x+27})/x

Lösung

\int \frac{3 x + 27}{x} d x

3 (3 x + 27 + ln 3 x + 27 - 3 - \frac{1}{2} ln 4 (x + 27)^{\frac{2}{3}} + 12 3 x + 27 + 36 - 3 arctan (\frac{2}{3 3} 3 x + 27 + \frac{1}{3})) + C

Schritte zur Lösung

\int \frac{3 x + 27}{x} d x

Wende U-Substitution an

= \int \frac{3 u ^{3}}{u ^{3} - 27} d u

Entferne die Konstante:

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

= 3 \cdot \int \frac{u ^{3}}{u ^{3} - 27} d u

Ermittle den Partialbruch von

\frac{u ^{3}}{u ^{3} - 27} : 1 + \frac{1}{u - 3} + \frac{- u - 6}{u ^{2} + 3 u + 9}

= 3 \cdot \int 1 + \frac{1}{u - 3} + \frac{- u - 6}{u ^{2} + 3 u + 9} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int 1 d u + \int \frac{1}{u - 3} d u + \int \frac{- u - 6}{u ^{2} + 3 u + 9} d u)

\int 1 d u = u

\int \frac{1}{u - 3} d u = ln ∣ u - 3 ∣

\int \frac{- u - 6}{u ^{2} + 3 u + 9} d u = - \frac{1}{2} ln 4 u^{2} + 12 u + 36 - 3 arctan (\frac{2}{3 3} u + \frac{1}{3})

= 3 (u + ln ∣ u - 3 ∣ - \frac{1}{2} ln 4 u^{2} + 12 u + 36 - 3 arctan (\frac{2}{3 3} u + \frac{1}{3}))

Setze in

u = 3 x + 27

ein

= 3 (3 x + 27 + ln 3 x + 27 - 3 - \frac{1}{2} ln 4 (3 x + 27)^{2} + 12 3 x + 27 + 36 - 3 arctan (\frac{2}{3 3} 3 x + 27 + \frac{1}{3}))

(3 x + 27)^{2} = (x + 27)^{\frac{2}{3}}

= 3 (3 x + 27 + ln 3 x + 27 - 3 - \frac{1}{2} ln 4 (x + 27)^{\frac{2}{3}} + 12 3 x + 27 + 36 - 3 arctan (\frac{2}{3 3} 3 x + 27 + \frac{1}{3}))

Füge eine Konstante zur Lösung hinzu

= 3 (3 x + 27 + ln 3 x + 27 - 3 - \frac{1}{2} ln 4 (x + 27)^{\frac{2}{3}} + 12 3 x + 27 + 36 - 3 arctan (\frac{2}{3 3} 3 x + 27 + \frac{1}{3})) + C

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

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

\int \frac{( 12 x + 1 )}{4 x - 3} d x

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

\int \frac{1}{( x ^{2} + 2 x + 10 ) ^{2}} d x