integral of (4x+1)/(2x^2+x)

Lösung

\int \frac{4 x + 1}{2 x ^{2} + x} d x

ln ∣ x ∣ + ln ∣ 2 x + 1 ∣ + C

Schritte zur Lösung

\int \frac{4 x + 1}{2 x ^{2} + x} d x

Ermittle den Partialbruch von

\frac{4 x + 1}{2 x ^{2} + x} : \frac{1}{x} + \frac{2}{2 x + 1}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ + ln ∣ 2 x + 1 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ x ∣ + ln ∣ 2 x + 1 ∣ + C

\frac{\partial}{\partial x} (x^{2} sin (11 t))

\frac{d}{d x} (\frac{x ^{4} - 5 x ^{3} + 8}{x ^{2}})

\frac{d}{d x} (e^{- x^{9}})

\int (tan^{2} (x) sec (x)) d x

x \to 1 lim (4 x + 3)