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

Lösung

integral

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

\int 4 d x = 4 x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{1}{( y ^{2} + 1 ) ^{2}}

in t e g r a l \frac{1}{2 sin ^{2} ( x )}

in t e g r a l \frac{x ^{2}}{x + 1}

in t e g r a l \frac{1}{2 ( x )}

in t e g r a l (x + 1) sin (x^{2} + 2 x + 3)