integral of (e^x)/3-4x^3+6/x

Lösung

\int \frac{e ^{x}}{3} - 4 x^{3} + \frac{6}{x} d x

\frac{1}{3} e^{x} - x^{4} + 6 ln ∣ x ∣ + C

Schritte zur Lösung

\int \frac{e ^{x}}{3} - 4 x^{3} + \frac{6}{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 \frac{e ^{x}}{3} d x - \int 4 x^{3} d x + \int \frac{6}{x} d x

\int \frac{e ^{x}}{3} d x = \frac{1}{3} e^{x}

\int 4 x^{3} d x = x^{4}

\int \frac{6}{x} d x = 6 ln ∣ x ∣

= \frac{1}{3} e^{x} - x^{4} + 6 ln ∣ x ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} e^{x} - x^{4} + 6 ln ∣ x ∣ + C

\frac{\partial}{\partial x} (ln (2 x^{2} + y^{2}))

n = 1 \sum \infty \frac{3}{n ^{2}}

(x y^{2} + x) d x + (y - x^{2} y) d y = 0

x \to 0 lim (x sin (\frac{4 π}{x}))

\int_{1}^{2} 27 x^{2} 4^{x^{3}} d x