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

Lösung

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

\frac{3}{2} ln ∣ 2 x - 1 ∣ - 2 ln ∣ x + 1 ∣ + C

Schritte zur Lösung

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

Ermittle den Partialbruch von

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

= \int \frac{3}{2 x - 1} - \frac{2}{x + 1} 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{3}{2 x - 1} d x - \int \frac{2}{x + 1} d x

\int \frac{3}{2 x - 1} d x = \frac{3}{2} ln ∣ 2 x - 1 ∣

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

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

Füge eine Konstante zur Lösung hinzu

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

z \to 2 lim (\frac{z ^{2} + 3}{i z})

t a y l or ln (1 + 7 x)

\frac{d ^{2}}{d x ^{2}} (4 x + 15)

\frac{\partial}{\partial x} (xx + yy)

\int_{0}^{π} x \cdot cos (x) d x