integral of (te^t+t)

Lösung

\int (t e^{t} + t) d t

e^{t} t - e^{t} + \frac{t ^{2}}{2} + C

Schritte zur Lösung

\int t e^{t} + t d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t e^{t} d t + \int t d t

\int t e^{t} d t = e^{t} t - e^{t}

\int t d t = \frac{t ^{2}}{2}

= e^{t} t - e^{t} + \frac{t ^{2}}{2}

Füge eine Konstante zur Lösung hinzu

= e^{t} t - e^{t} + \frac{t ^{2}}{2} + C