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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

= \int 3 x - 2 + \frac{1}{x} - \frac{5}{x ^{2}} 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 3 x d x - \int 2 d x + \int \frac{1}{x} d x - \int \frac{5}{x ^{2}} d x

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

\int 2 d x = 2 x

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

t an g e n t f (x) = \frac{e ^{x}}{x ^{2}}, at x = 1

n = 1 \sum \infty \frac{3}{5 ^{n}}

\int \frac{3 x ^{5}}{1 - 12 x ^{6}} d x

\int 9 x e^{- x^{2}} d x

\int ln (x + 5) d x