integral of 4xe^{2/3 x}

Lösung

\int 4 x e^{\frac{2}{3} x} d x

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

Schritte zur Lösung

\int 4 x e^{\frac{2}{3} x} d x

Entferne die Konstante:

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

= 4 \cdot \int x e^{\frac{2}{3} x} d x

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= 4 \cdot \int x e^{\frac{2 x}{3}} d x

Wende U-Substitution an

= 4 \cdot \int \frac{9 e ^{u} u}{4} d u

Entferne die Konstante:

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

= 4 \cdot \frac{9}{4} \cdot \int e^{u} u d u

Wende die partielle Integration an

= 4 \cdot \frac{9}{4} (e^{u} u - \int e^{u} d u)

\int e^{u} d u = e^{u}

= 4 \cdot \frac{9}{4} (e^{u} u - e^{u})

Setze in

u = \frac{2 x}{3}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

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

\frac{d ^{2}}{d x ^{2}} ((2 x^{2} + 7)^{5})

\frac{d}{d x} (I n, x^{2})

\frac{d P}{d t} = k P

f (x) = 4 tan (x)