integral of (1-t)(7+t^2)

Lösung

\int (1 - t) (7 + t^{2}) d t

7 t + \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} - \frac{t ^{4}}{4} + C

Schritte zur Lösung

\int (1 - t) (7 + t^{2}) d t

Multipliziere aus

(1 - t) (7 + t^{2}) : 7 + t^{2} - 7 t - t^{3}

= \int 7 + t^{2} - 7 t - t^{3} d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 7 d t + \int t^{2} d t - \int 7 t d t - \int t^{3} d t

\int 7 d t = 7 t

\int t^{2} d t = \frac{t ^{3}}{3}

\int 7 t d t = \frac{7 t ^{2}}{2}

\int t^{3} d t = \frac{t ^{4}}{4}

= 7 t + \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} - \frac{t ^{4}}{4}

Füge eine Konstante zur Lösung hinzu

= 7 t + \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} - \frac{t ^{4}}{4} + C

x \to 1 + lim (2 - x)

\int \frac{( ( ln ( x ) ) ^{2} )}{x} d x

\frac{d}{d x} (\frac{x ^{7}}{7})

s l o p e f (x) = - 0.04 x^{4} + 4

f (x) = arctan (e^{2 x})