integral from 1 to 3 of (2t+3)/(t+1)

Lösung

\int_{1}^{3} \frac{2 t + 3}{t + 1} d t

4 + ln (2)

Dezimale

4.69314 \dots

Schritte zur Lösung

\int_{1}^{3} \frac{2 t + 3}{t + 1} d t

Wende U-Substitution an

= \int_{2}^{4} \frac{2 u + 1}{u} d u

Multipliziere aus

\frac{2 u + 1}{u} : 2 + \frac{1}{u}

= \int_{2}^{4} 2 + \frac{1}{u} d u

Wende die Summenregel an:

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

= \int_{2}^{4} 2 d u + \int_{2}^{4} \frac{1}{u} d u

\int_{2}^{4} 2 d u = 4

\int_{2}^{4} \frac{1}{u} d u = ln (2)

= 4 + ln (2)