integral of \sqrt[7]{x^3}(1+sqrt(x))

Lösung

\int 7 x^{3} (1 + x) d x

\frac{1}{3} (\frac{21}{10} x^{\frac{10}{7}} + \frac{14}{9} x^{\frac{27}{14}}) + C

Schritte zur Lösung

\int 7 x^{3} (1 + x) d x

Wende U-Substitution an

= \int \frac{1 + 6 u}{3 u ^{\frac{11}{21}}} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int \frac{1 + 6 u}{u ^{\frac{11}{21}}} d u

Multipliziere aus

\frac{1 + 6 u}{u ^{\frac{11}{21}}} : \frac{1}{u ^{\frac{11}{21}}} + \frac{1}{u ^{\frac{5}{14}}}

= \frac{1}{3} \cdot \int \frac{1}{u ^{\frac{11}{21}}} + \frac{1}{u ^{\frac{5}{14}}} 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}{3} (\int \frac{1}{u ^{\frac{11}{21}}} d u + \int \frac{1}{u ^{\frac{5}{14}}} d u)

\int \frac{1}{u ^{\frac{11}{21}}} d u = \frac{21}{10} u^{\frac{10}{21}}

\int \frac{1}{u ^{\frac{5}{14}}} d u = \frac{14}{9} u^{\frac{9}{14}}

= \frac{1}{3} (\frac{21}{10} u^{\frac{10}{21}} + \frac{14}{9} u^{\frac{9}{14}})

Setze in

u = x^{3}

ein

= \frac{1}{3} (\frac{21}{10} (x^{3})^{\frac{10}{21}} + \frac{14}{9} (x^{3})^{\frac{9}{14}})

Vereinfache

\frac{1}{3} (\frac{21}{10} (x^{3})^{\frac{10}{21}} + \frac{14}{9} (x^{3})^{\frac{9}{14}}) : \frac{1}{3} (\frac{21}{10} x^{\frac{10}{7}} + \frac{14}{9} x^{\frac{27}{14}})

= \frac{1}{3} (\frac{21}{10} x^{\frac{10}{7}} + \frac{14}{9} x^{\frac{27}{14}})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (\frac{21}{10} x^{\frac{10}{7}} + \frac{14}{9} x^{\frac{27}{14}}) + C

x \to 1.9 lim (- x^{2} + 5 x - 2)

x \to - 5 lim (\frac{- 6 x}{x + 5})

\frac{d y}{d t} = - 3 (y - 2)

\frac{\partial}{\partial x} (3 x + 2 x y^{2})

d er i v a t i v e y = arccos (\frac{11}{x})