integral of (u^5-3u^4-u^2+2/7)

Lösung

\int (u^{5} - 3 u^{4} - u^{2} + \frac{2}{7}) d u

\frac{u ^{6}}{6} - \frac{3 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{2}{7} u + C

Schritte zur Lösung

\int u^{5} - 3 u^{4} - u^{2} + \frac{2}{7} d u

Wende die Summenregel an:

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

= \int u^{5} d u - \int 3 u^{4} d u - \int u^{2} d u + \int \frac{2}{7} d u

\int u^{5} d u = \frac{u ^{6}}{6}

\int 3 u^{4} d u = \frac{3 u ^{5}}{5}

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

\int \frac{2}{7} d u = \frac{2}{7} u

= \frac{u ^{6}}{6} - \frac{3 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{2}{7} u

Füge eine Konstante zur Lösung hinzu

= \frac{u ^{6}}{6} - \frac{3 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{2}{7} u + C

\frac{d}{d x} (2 x sin (x) + cos (x) x^{2})

x \to \infty lim (3 \cdot x)

\int \frac{17}{x ln ( 3 x )} d x

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

t an g e n t 4 x (3 - 2 x)^{5}, at x = 2