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Beliebt Rechnen >

integral of (5x^2-4)^2x^3

Lösung

\int (5 x^{2} - 4)^{2} x^{3} d x

Lösung

\frac{1}{2} (\frac{25 x ^{8}}{4} - \frac{40 x ^{6}}{3} + 8 x^{4}) + C

Schritte zur Lösung

\int (5 x^{2} - 4)^{2} x^{3} d x

Wende U-Substitution an

= \int \frac{u ( 5 u - 4 ) ^{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 (5 u - 4)^{2} d u

Faktorisiere

5 u - 4 : - (- 5 u + 4)

= \frac{1}{2} \cdot \int u (- (- 5 u + 4))^{2} d u

(- (- 5 u + 4))^{2} = (- 5 u + 4)^{2}

= \frac{1}{2} \cdot \int u (- 5 u + 4)^{2} d u

Multipliziere aus

u (- 5 u + 4)^{2} : 25 u^{3} - 40 u^{2} + 16 u

= \frac{1}{2} \cdot \int 25 u^{3} - 40 u^{2} + 16 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}{2} (\int 25 u^{3} d u - \int 40 u^{2} d u + \int 16 u d u)

\int 25 u^{3} d u = \frac{25 u ^{4}}{4}

\int 40 u^{2} d u = \frac{40 u ^{3}}{3}

\int 16 u d u = 8 u^{2}

= \frac{1}{2} (\frac{25 u ^{4}}{4} - \frac{40 u ^{3}}{3} + 8 u^{2})

Setze in

u = x^{2}

ein

= \frac{1}{2} (\frac{25 ( x ^{2} ) ^{4}}{4} - \frac{40 ( x ^{2} ) ^{3}}{3} + 8 (x^{2})^{2})

Vereinfache

\frac{1}{2} (\frac{25 ( x ^{2} ) ^{4}}{4} - \frac{40 ( x ^{2} ) ^{3}}{3} + 8 (x^{2})^{2}) : \frac{1}{2} (\frac{25 x ^{8}}{4} - \frac{40 x ^{6}}{3} + 8 x^{4})

= \frac{1}{2} (\frac{25 x ^{8}}{4} - \frac{40 x ^{6}}{3} + 8 x^{4})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (\frac{25 x ^{8}}{4} - \frac{40 x ^{6}}{3} + 8 x^{4}) + C

Graph

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