integral of (5x^2+2xe^x-1)/x

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

Füge eine Konstante zur Lösung hinzu

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

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