integral of 1/((x-1)(x^2+9))

Lösung

\int \frac{1}{( x - 1 ) ( x ^{2} + 9 )} d x

\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})) + C

Schritte zur Lösung

\int \frac{1}{( x - 1 ) ( x ^{2} + 9 )} d x

Ermittle den Partialbruch von

\frac{1}{( x - 1 ) ( x ^{2} + 9 )} : \frac{1}{10 ( x - 1 )} + \frac{- x - 1}{10 ( x ^{2} + 9 )}

= \int \frac{1}{10 ( x - 1 )} + \frac{- x - 1}{10 ( x ^{2} + 9 )} 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}{10 ( x - 1 )} d x + \int \frac{- x - 1}{10 ( x ^{2} + 9 )} d x

\int \frac{1}{10 ( x - 1 )} d x = \frac{1}{10} ln ∣ x - 1 ∣

\int \frac{- x - 1}{10 ( x ^{2} + 9 )} d x = \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))

= \frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})) + C

\int \frac{1}{9 - 4 x} d x

d er i v a t i v e y = \frac{- 12}{4 x}

2 x + y + (x + 2 y) y^{^{'}} = 0

\int - 2 x^{2} + 6 x d x

d er i v a t i v e (cos^{2} (s) - 11 s^{2})^{2}