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Beliebt Rechnen >

integral of (x^2+2x)e^{3x+2}

Lösung

\int (x^{2} + 2 x) e^{3 x + 2} d x

Lösung

\frac{e ^{2}}{27} (9 e^{3 x} x^{2} + 4 (3 e^{3 x} x - e^{3 x})) + C

Schritte zur Lösung

\int (x^{2} + 2 x) e^{3 x + 2} d x

(x^{2} + 2 x) e^{3 x + 2} = e^{2} (x^{2} + 2 x) e^{3 x}

= \int e^{2} (x^{2} + 2 x) e^{3 x} d x

Entferne die Konstante:

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

= e^{2} \cdot \int (x^{2} + 2 x) e^{3 x} d x

Wende U-Substitution an

= e^{2} \cdot \int \frac{1}{3} e^{u} (\frac{u ^{2}}{9} + \frac{2 u}{3}) d u

Entferne die Konstante:

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

= e^{2} \frac{1}{3} \cdot \int e^{u} (\frac{u ^{2}}{9} + \frac{2 u}{3}) d u

Multipliziere aus

e^{u} (\frac{u ^{2}}{9} + \frac{2 u}{3}) : \frac{e ^{u} u ^{2}}{9} + \frac{2 e ^{u} u}{3}

= e^{2} \frac{1}{3} \cdot \int \frac{e ^{u} u ^{2}}{9} + \frac{2 e ^{u} u}{3} 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

= e^{2} \frac{1}{3} (\int \frac{e ^{u} u ^{2}}{9} d u + \int \frac{2 e ^{u} u}{3} d u)

\int \frac{e ^{u} u ^{2}}{9} d u = \frac{1}{9} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

\int \frac{2 e ^{u} u}{3} d u = \frac{2}{3} (e^{u} u - e^{u})

= e^{2} \frac{1}{3} (\frac{1}{9} (u^{2} e^{u} - 2 (e^{u} u - e^{u})) + \frac{2}{3} (e^{u} u - e^{u}))

Setze in

u = 3 x

ein

= e^{2} \frac{1}{3} (\frac{1}{9} ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x})) + \frac{2}{3} (e^{3 x} \cdot 3 x - e^{3 x}))

Vereinfache

e^{2} \frac{1}{3} (\frac{1}{9} ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x})) + \frac{2}{3} (e^{3 x} \cdot 3 x - e^{3 x})) : \frac{e ^{2}}{27} (9 e^{3 x} x^{2} + 4 (3 e^{3 x} x - e^{3 x}))

= \frac{e ^{2}}{27} (9 e^{3 x} x^{2} + 4 (3 e^{3 x} x - e^{3 x}))

Füge eine Konstante zur Lösung hinzu

= \frac{e ^{2}}{27} (9 e^{3 x} x^{2} + 4 (3 e^{3 x} x - e^{3 x})) + C

Graph

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