integral of (e-x-e^3)(ex+e-5)

Lösung

\int (e - x - e^{3}) (e x + e - 5) d x

\frac{e ^{2} x ^{2}}{2} + e^{2} x - 5 e x - \frac{e x ^{3}}{3} - \frac{e x ^{2}}{2} + \frac{5 x ^{2}}{2} - \frac{e ^{4} x ^{2}}{2} - e^{4} x + 5 e^{3} x + C

Schritte zur Lösung

\int (e - x - e^{3}) (e x + e - 5) d x

Multipliziere aus

(e - x - e^{3}) (e x + e - 5) : e^{2} x + e^{2} - 5 e - e x^{2} - e x + 5 x - e^{4} x - e^{4} + 5 e^{3}

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

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

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

\int 5 e d x = 5 e x

\int e x^{2} d x = \frac{e x ^{3}}{3}

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

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

\int e^{4} x d x = \frac{e ^{4} x ^{2}}{2}

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

\int 5 e^{3} d x = 5 e^{3} x

= \frac{e ^{2} x ^{2}}{2} + e^{2} x - 5 e x - \frac{e x ^{3}}{3} - \frac{e x ^{2}}{2} + \frac{5 x ^{2}}{2} - \frac{e ^{4} x ^{2}}{2} - e^{4} x + 5 e^{3} x

Füge eine Konstante zur Lösung hinzu

= \frac{e ^{2} x ^{2}}{2} + e^{2} x - 5 e x - \frac{e x ^{3}}{3} - \frac{e x ^{2}}{2} + \frac{5 x ^{2}}{2} - \frac{e ^{4} x ^{2}}{2} - e^{4} x + 5 e^{3} x + C

\int_{9}^{x} 1 + 9 t d t

\frac{d}{d x} (lo g_{e} (x^{3}))

\int_{0}^{2} \frac{1}{8} (x + y) d y

f (x) = e^{2 x} - 1

t an g e n t y = x^{3} - 2 x^{2} + 5, at x = 2