integral of (4x^3-2x+1)/x

Lösung

\int \frac{4 x ^{3} - 2 x + 1}{x} d x

\frac{4 x ^{3}}{3} - 2 x + ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{4 x ^{3} - 2 x + 1}{x} d x

Multipliziere aus

\frac{4 x ^{3} - 2 x + 1}{x} : 4 x^{2} - 2 + \frac{1}{x}

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

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 2 d x = 2 x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{4 x ^{3}}{3} - 2 x + ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{4 x ^{3}}{3} - 2 x + ln ∣ x ∣ + C

t an g e n t y = 3 x^{3} - x^{2} + 1, (2, 21)

x \to \infty lim (2 x^{2} + 1)

\frac{d}{d x} (e^{x} + e)

\int \frac{sin ^{2} ( x )}{cos ^{3} ( x )} d x

a re a y = \frac{26}{5} - \frac{1}{x}, y = x