integral from 1 to 2 of (t+1/t)^2

Lösung

\int_{1}^{2} (t + \frac{1}{t})^{2} d t

\frac{29}{6}

Dezimale

4.83333 \dots

Schritte zur Lösung

\int_{1}^{2} (t + \frac{1}{t})^{2} d t

Multipliziere aus

(t + \frac{1}{t})^{2} : t^{2} + 2 + \frac{1}{t ^{2}}

= \int_{1}^{2} t^{2} + 2 + \frac{1}{t ^{2}} 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}^{2} t^{2} d t + \int_{1}^{2} 2 d t + \int_{1}^{2} \frac{1}{t ^{2}} d t

\int_{1}^{2} t^{2} d t = \frac{7}{3}

\int_{1}^{2} 2 d t = 2

\int_{1}^{2} \frac{1}{t ^{2}} d t = \frac{1}{2}

= \frac{7}{3} + 2 + \frac{1}{2}

Vereinfache

\frac{7}{3} + 2 + \frac{1}{2} : \frac{29}{6}

= \frac{29}{6}

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