integral of x(x-1)^4

Lösung

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

\frac{x ^{6}}{6} - \frac{4 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2} + C

Schritte zur Lösung

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

Multipliziere aus

x (x - 1)^{4} : x^{5} - 4 x^{4} + 6 x^{3} - 4 x^{2} + x

= \int x^{5} - 4 x^{4} + 6 x^{3} - 4 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 x^{5} d x - \int 4 x^{4} d x + \int 6 x^{3} d x - \int 4 x^{2} d x + \int x d x

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

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

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

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

= \frac{x ^{6}}{6} - \frac{4 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= \frac{x ^{6}}{6} - \frac{4 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2} + C

\int \frac{9}{x ^{2} x ^{2} + 16} d x

d er i v a t i v e tan^{2} (5 θ)

n = 1 \sum \infty 5 (- 1)^{n} n x^{n}

l a pl a ce t r an s f or m e^{- 4}

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