integral of x(4x-1)^4

Lösung

\int x (4 x - 1)^{4} d x

\frac{128 x ^{6}}{3} - \frac{256 x ^{5}}{5} + 24 x^{4} - \frac{16 x ^{3}}{3} + \frac{x ^{2}}{2} + C

Schritte zur Lösung

\int x (4 x - 1)^{4} d x

Multipliziere aus

x (4 x - 1)^{4} : 256 x^{5} - 256 x^{4} + 96 x^{3} - 16 x^{2} + x

= \int 256 x^{5} - 256 x^{4} + 96 x^{3} - 16 x^{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 256 x^{5} d x - \int 256 x^{4} d x + \int 96 x^{3} d x - \int 16 x^{2} d x + \int x d x

\int 256 x^{5} d x = \frac{128 x ^{6}}{3}

\int 256 x^{4} d x = \frac{256 x ^{5}}{5}

\int 96 x^{3} d x = 24 x^{4}

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

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

= \frac{128 x ^{6}}{3} - \frac{256 x ^{5}}{5} + 24 x^{4} - \frac{16 x ^{3}}{3} + \frac{x ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= \frac{128 x ^{6}}{3} - \frac{256 x ^{5}}{5} + 24 x^{4} - \frac{16 x ^{3}}{3} + \frac{x ^{2}}{2} + C

d er i v a t i v e y = ln (8 x)

y^{^{''}} - 7 y^{^{'}} = 5 t

x \to 0 lim (\frac{x}{sin ( 4 x )})

l a pl a ce t r an s f or m sin (t - π)

t an g e n t f (x) = x^{2}, at x = 0.2