integral of 8\sqrt[3]{x}+(x+1)e^{x^2+2x}

Lösung

\int 8 3 x + (x + 1) e^{x^{2} + 2 x} d x

6 x^{\frac{4}{3}} + \frac{1}{2} e^{x^{2} + 2 x} + C

Schritte zur Lösung

\int 8 3 x + (x + 1) e^{x^{2} + 2 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 8 3 x d x + \int (x + 1) e^{x^{2} + 2 x} d x

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

\int (x + 1) e^{x^{2} + 2 x} d x = \frac{1}{2} e^{x^{2} + 2 x}

= 6 x^{\frac{4}{3}} + \frac{1}{2} e^{x^{2} + 2 x}

Füge eine Konstante zur Lösung hinzu

= 6 x^{\frac{4}{3}} + \frac{1}{2} e^{x^{2} + 2 x} + C

\frac{d}{d x} (1 - x^{2} arccos (x))

\int cos (e^{t}) d t

\int_{0}^{5} (5 x - x^{2}) d x

\int 9 d x

t an g e n t \frac{4 x}{x + 1}, at (3.3)