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

Lösung

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

x^{2} + ln ∣ x ∣ + C

Schritte zur Lösung

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

Multipliziere aus

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

= \int 2 x + \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 2 x d x + \int \frac{1}{x} d x

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

\int \frac{1}{x} d x = ln ∣ x ∣

= x^{2} + ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= x^{2} + ln ∣ x ∣ + C

x \to 1 lim (\frac{x ^{4} - 1}{x ^{2}})

d er i v a t i v e f (x) = (2 x^{2} - 4)^{- 1}

x \to 8 - lim (\frac{x}{x ^{2} - 64})

\frac{d}{d x} (\frac{- 2 x ^{2} + 9}{9 - x ^{2}})

y^{^{'}} = (y + 2 t)^{2}, y (0) = 3