integral of ((x+1))/(x^3+x^2+6x)

Lösung

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

\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln 4 x^{2} + 4 x + 24 + \frac{11}{6 23} arctan (\frac{1}{23} (2 x + 1)) + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

\frac{x + 1}{x ^{3} + x ^{2} + 6 x} : \frac{1}{6 x} + \frac{- x + 5}{6 ( x ^{2} + x + 6 )}

= \int \frac{1}{6 x} + \frac{- x + 5}{6 ( x ^{2} + x + 6 )} 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}{6 x} d x + \int \frac{- x + 5}{6 ( x ^{2} + x + 6 )} d x

\int \frac{1}{6 x} d x = \frac{1}{6} ln ∣ x ∣

\int \frac{- x + 5}{6 ( x ^{2} + x + 6 )} d x = \frac{1}{12 23} (- 23 ln 4 x^{2} + 4 x + 24 + 22 arctan (\frac{2 x + 1}{23}))

= \frac{1}{6} ln ∣ x ∣ + \frac{1}{12 23} (- 23 ln 4 x^{2} + 4 x + 24 + 22 arctan (\frac{2 x + 1}{23}))

Vereinfache

\frac{1}{6} ln ∣ x ∣ + \frac{1}{12 23} (- 23 ln 4 x^{2} + 4 x + 24 + 22 arctan (\frac{2 x + 1}{23})) : \frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln 4 x^{2} + 4 x + 24 + \frac{11}{6 23} arctan (\frac{1}{23} (2 x + 1))

= \frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln 4 x^{2} + 4 x + 24 + \frac{11}{6 23} arctan (\frac{1}{23} (2 x + 1))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln 4 x^{2} + 4 x + 24 + \frac{11}{6 23} arctan (\frac{1}{23} (2 x + 1)) + C

x^{^{'}} = 4 x - x^{3}

\frac{d}{d x} (\frac{x - 5}{x + 6})

\frac{d}{d x} (b x)

y^{^{'}} = t + 3

\frac{\partial}{\partial x} (\frac{( 3 x + 5 )}{6 y + 1})