integral of (3\sqrt[4]{x^3}-8x^5+6e^x-2)

Lösung

\int (3 4 x^{3} - 8 x^{5} + 6 e^{x} - 2) d x

\frac{12}{7} x^{\frac{7}{4}} - \frac{4 x ^{6}}{3} + 6 e^{x} - 2 x + C

Schritte zur Lösung

\int 3 4 x^{3} - 8 x^{5} + 6 e^{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 3 4 x^{3} d x - \int 8 x^{5} d x + \int 6 e^{x} d x - \int 2 d x

\int 3 4 x^{3} d x = \frac{12}{7} x^{\frac{7}{4}}

\int 8 x^{5} d x = \frac{4 x ^{6}}{3}

\int 6 e^{x} d x = 6 e^{x}

\int 2 d x = 2 x

= \frac{12}{7} x^{\frac{7}{4}} - \frac{4 x ^{6}}{3} + 6 e^{x} - 2 x

Füge eine Konstante zur Lösung hinzu

= \frac{12}{7} x^{\frac{7}{4}} - \frac{4 x ^{6}}{3} + 6 e^{x} - 2 x + C

\frac{d y}{d x} = y (3 - y)

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

\frac{d}{d x} (+ 2 x)

t an g e n t f (x) = 6 x, (9, 18)

f (x) = tan^{3} (x)