integral of 3x^5+4x

Lösung

\int 3 x^{5} + 4 x d x

\frac{x ^{6}}{2} + 2 x^{2} + C

Schritte zur Lösung

\int 3 x^{5} + 4 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 3 x^{5} d x + \int 4 x d x

\int 3 x^{5} d x = \frac{x ^{6}}{2}

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

= \frac{x ^{6}}{2} + 2 x^{2}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{6}}{2} + 2 x^{2} + C

d er i v a t i v e f (x) = \frac{4}{x + 1}

\frac{d}{d x} (\frac{( 2 x + 1 ) ^{5}}{x ^{2} + 1})

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

\frac{d}{d z} (ln (\frac{c ^{2} - z ^{2}}{c ^{2} + z ^{2}}))

y^{^{''}} + 4 y^{^{'}} + 6 y = 1 + e^{- t}, y (0) = 0, y^{^{'}} (0) = 0