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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

\int 3 d x = 3 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

Füge eine Konstante zur Lösung hinzu

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