integral x^3(x^2+4)^{1/2}

Lösung

integral

x^{3} (x^{2} + 4)^{\frac{1}{2}}

\frac{1}{15} (3 x^{2} - 8) (x^{2} + 4)^{\frac{3}{2}} + C

Schritte zur Lösung

\int x^{3} (x^{2} + 4)^{\frac{1}{2}} d x

Wende U-Substitution an

= \int \frac{( u - 4 ) u ^{\frac{1}{2}}}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int (u - 4) u^{\frac{1}{2}} d u

Multipliziere aus

(u - 4) u^{\frac{1}{2}} : u^{\frac{3}{2}} - 4 u^{\frac{1}{2}}

= \frac{1}{2} \cdot \int u^{\frac{3}{2}} - 4 u^{\frac{1}{2}} 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}{2} (\int u^{\frac{3}{2}} d u - \int 4 u^{\frac{1}{2}} d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 4 u^{\frac{1}{2}} d u = \frac{8}{3} u^{\frac{3}{2}}

= \frac{1}{2} (\frac{2}{5} u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}})

Setze in

u = x^{2} + 4

ein

= \frac{1}{2} (\frac{2}{5} (x^{2} + 4)^{\frac{5}{2}} - \frac{8}{3} (x^{2} + 4)^{\frac{3}{2}})

Vereinfache

\frac{1}{2} (\frac{2}{5} (x^{2} + 4)^{\frac{5}{2}} - \frac{8}{3} (x^{2} + 4)^{\frac{3}{2}}) : \frac{1}{15} (3 x^{2} - 8) (x^{2} + 4)^{\frac{3}{2}}

= \frac{1}{15} (3 x^{2} - 8) (x^{2} + 4)^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{15} (3 x^{2} - 8) (x^{2} + 4)^{\frac{3}{2}} + C

in t e g r a l \frac{sin ^{7} ( x )}{cos ^{2} ( x )}

in t e g r a l x^{2} - 2 x + 1

in t e g r a l \frac{1}{( 1 + r ^{2} ) ^{2}}

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

in t e g r a l \frac{1}{( 2 + sin ( 3 x ) )}