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

Lösung

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

\frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln 4 x^{2} + 4 x + 20 - \frac{1}{5 19} arctan (\frac{1}{19} (2 x + 1)) + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

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

\int \frac{1}{5 x} d x = \frac{1}{5} ln ∣ x ∣

\int \frac{- x - 1}{5 ( x ^{2} + x + 5 )} d x = \frac{1}{10 19} (- 19 ln 4 x^{2} + 4 x + 20 - 2 arctan (\frac{2 x + 1}{19}))

= \frac{1}{5} ln ∣ x ∣ + \frac{1}{10 19} (- 19 ln 4 x^{2} + 4 x + 20 - 2 arctan (\frac{2 x + 1}{19}))

Vereinfache

\frac{1}{5} ln ∣ x ∣ + \frac{1}{10 19} (- 19 ln 4 x^{2} + 4 x + 20 - 2 arctan (\frac{2 x + 1}{19})) : \frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln 4 x^{2} + 4 x + 20 - \frac{1}{5 19} arctan (\frac{1}{19} (2 x + 1))

= \frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln 4 x^{2} + 4 x + 20 - \frac{1}{5 19} arctan (\frac{1}{19} (2 x + 1))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} ln ∣ x ∣ - \frac{1}{10} ln 4 x^{2} + 4 x + 20 - \frac{1}{5 19} arctan (\frac{1}{19} (2 x + 1)) + C

\frac{\partial}{\partial x} ((6 - x^{2} y^{2})^{2})

x \to 4 + lim (\frac{4 + x}{4 - x})

h \to 0 lim (6 x + 1 + 3 h)

in v erse l a pl a ce {\frac{1}{s ^{2} + s - 20}}

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