integral of x(x+6)^7

Lösung

\int x (x + 6)^{7} d x

\frac{1}{36} (x + 6)^{8} (4 x - 3) + C

Schritte zur Lösung

\int x (x + 6)^{7} d x

Wende U-Substitution an

= \int \frac{7 u ( 7 u - 6 )}{7} d u

Entferne die Konstante:

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

= \frac{1}{7} \cdot \int 7 u (7 u - 6) d u

Multipliziere aus

7 u (7 u - 6) : u^{\frac{2}{7}} - 6 7 u

= \frac{1}{7} \cdot \int u^{\frac{2}{7}} - 6 7 u 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

= \frac{1}{7} (\int u^{\frac{2}{7}} d u - \int 6 7 u d u)

\int u^{\frac{2}{7}} d u = \frac{7}{9} u^{\frac{9}{7}}

\int 6 7 u d u = \frac{21}{4} u^{\frac{8}{7}}

= \frac{1}{7} (\frac{7}{9} u^{\frac{9}{7}} - \frac{21}{4} u^{\frac{8}{7}})

Setze in

u = (x + 6)^{7}

ein

= \frac{1}{7} (\frac{7}{9} ((x + 6)^{7})^{\frac{9}{7}} - \frac{21}{4} ((x + 6)^{7})^{\frac{8}{7}})

Vereinfache

\frac{1}{7} (\frac{7}{9} ((x + 6)^{7})^{\frac{9}{7}} - \frac{21}{4} ((x + 6)^{7})^{\frac{8}{7}}) : \frac{1}{36} (x + 6)^{8} (4 x - 3)

= \frac{1}{36} (x + 6)^{8} (4 x - 3)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{36} (x + 6)^{8} (4 x - 3) + C

\frac{\partial}{\partial y} (cos (x) + x cos (y) - y)

n = 0 \sum \infty (\frac{2}{4})^{n}

\int 5 x + 12 d x

n = 0 \sum \infty n \cdot (\frac{1}{6})^{n}

x y^{^{'}} + 2 y - 2 x^{2} y^{2} = 0