integral 6(2-x)^5

Lösung

integral

6 (2 - x)^{5}

192 x - 240 x^{2} + 160 x^{3} - 60 x^{4} + 12 x^{5} - x^{6} + C

Schritte zur Lösung

\int 6 (2 - x)^{5} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int (2 - x)^{5} d x

Multipliziere aus

(2 - x)^{5} : 32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - x^{5}

= 6 \cdot \int 32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - x^{5} 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

= 6 (\int 32 d x - \int 80 x d x + \int 80 x^{2} d x - \int 40 x^{3} d x + \int 10 x^{4} d x - \int x^{5} d x)

\int 32 d x = 32 x

\int 80 x d x = 40 x^{2}

\int 80 x^{2} d x = \frac{80 x ^{3}}{3}

\int 40 x^{3} d x = 10 x^{4}

\int 10 x^{4} d x = 2 x^{5}

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

= 6 (32 x - 40 x^{2} + \frac{80 x ^{3}}{3} - 10 x^{4} + 2 x^{5} - \frac{x ^{6}}{6})

Vereinfache

6 (32 x - 40 x^{2} + \frac{80 x ^{3}}{3} - 10 x^{4} + 2 x^{5} - \frac{x ^{6}}{6}) : 192 x - 240 x^{2} + 160 x^{3} - 60 x^{4} + 12 x^{5} - x^{6}

= 192 x - 240 x^{2} + 160 x^{3} - 60 x^{4} + 12 x^{5} - x^{6}

Füge eine Konstante zur Lösung hinzu

= 192 x - 240 x^{2} + 160 x^{3} - 60 x^{4} + 12 x^{5} - x^{6} + C

in t e g r a l 4 x^{2} + 5

in t e g r a l \frac{3 x}{( 6 - 3 x ) ^{\frac{1}{2}}}

in t e g r a l \frac{x}{1 - x}

in t e g r a l (1 - t^{2})^{3}

in t e g r a l \frac{5 x ^{4}}{x ^{5} + 9}