integral from 1 to e of (t^4+1)/t

Lösung

\int_{1}^{e} \frac{t ^{4} + 1}{t} d t

\frac{e ^{4} - 1}{4} + 1

Dezimale

14.39953 \dots

Schritte zur Lösung

\int_{1}^{e} \frac{t ^{4} + 1}{t} d t

Multipliziere aus

\frac{t ^{4} + 1}{t} : t^{3} + \frac{1}{t}

= \int_{1}^{e} t^{3} + \frac{1}{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_{1}^{e} t^{3} d t + \int_{1}^{e} \frac{1}{t} d t

\int_{1}^{e} t^{3} d t = \frac{e ^{4} - 1}{4}

\int_{1}^{e} \frac{1}{t} d t = 1

= \frac{e ^{4} - 1}{4} + 1