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{3}{10} ln ∣ x - 2 ∣ - \frac{2}{15} ln ∣ x + 3 ∣ + 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{3}{10 ( x - 2 )} - \frac{2}{15 ( x + 3 )}

= \int - \frac{1}{6 x} + \frac{3}{10 ( x - 2 )} - \frac{2}{15 ( x + 3 )} 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{3}{10 ( x - 2 )} d x - \int \frac{2}{15 ( x + 3 )} d x

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

\int \frac{3}{10 ( x - 2 )} d x = \frac{3}{10} ln ∣ x - 2 ∣

\int \frac{2}{15 ( x + 3 )} d x = \frac{2}{15} ln ∣ x + 3 ∣

= - \frac{1}{6} ln ∣ x ∣ + \frac{3}{10} ln ∣ x - 2 ∣ - \frac{2}{15} ln ∣ x + 3 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{6} ln ∣ x ∣ + \frac{3}{10} ln ∣ x - 2 ∣ - \frac{2}{15} ln ∣ x + 3 ∣ + C

t an g e n t y = x^{2}, at x = - 0.5

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

d er i v a t i v e y = (4 x^{6} + 7)^{5}

d er i v a t i v e 9 x - 6 x

\frac{\partial}{\partial x} (2 cos (2 x))