integral of (t+4)e^{t^2+8t+1}

Lösung

\int (t + 4) e^{t^{2} + 8 t + 1} d t

\frac{1}{2} e^{1 + t^{2} + 8 t} + C

Schritte zur Lösung

\int (t + 4) e^{t^{2} + 8 t + 1} d t

(t + 4) e^{t^{2} + 8 t + 1} = e^{1} (t + 4) e^{t^{2} + 8 t}

= \int e^{1} (t + 4) e^{t^{2} + 8 t} d t

Entferne die Konstante:

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

= e^{1} \cdot \int (t + 4) e^{t^{2} + 8 t} d t

Vereinfache

= e \cdot \int (t + 4) e^{t^{2} + 8 t} d t

Wende U-Substitution an

= e \cdot \int \frac{e ^{u}}{2} d u

Entferne die Konstante:

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

= e \frac{1}{2} \cdot \int e^{u} d u

Nutze das gemeinsame Integral :

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

= e \frac{1}{2} e^{u}

Setze in

u = t^{2} + 8 t

ein

= e \frac{1}{2} e^{t^{2} + 8 t}

Wende Exponentenregel an:

a^{b} \cdot a^{c} = a^{b + c}

e e^{t^{2} + 8 t} = e^{1 + t^{2} + 8 t}

= \frac{1}{2} e^{1 + t^{2} + 8 t}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} e^{1 + t^{2} + 8 t} + C

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