integral of (2x^3-5x^2+x-6)/x

Lösung

\int \frac{2 x ^{3} - 5 x ^{2} + x - 6}{x} d x

\frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x - 6 ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{2 x ^{3} - 5 x ^{2} + x - 6}{x} d x

Multipliziere aus

\frac{2 x ^{3} - 5 x ^{2} + x - 6}{x} : 2 x^{2} - 5 x + 1 - \frac{6}{x}

= \int 2 x^{2} - 5 x + 1 - \frac{6}{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 2 x^{2} d x - \int 5 x d x + \int 1 d x - \int \frac{6}{x} d x

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

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 1 d x = x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

= \frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x - 6 ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} + x - 6 ln ∣ x ∣ + C

x \to 0 lim (\frac{16}{x ^{4}})

d er i v a t i v e \frac{e ^{y} + e ^{- y}}{2}

\int e^{y^{2}} d y

\int \frac{10}{8 x - 4 x ^{2}} d x

\int_{\frac{1}{2}}^{\frac{e}{2}} ln (2 x) d x