integral from 0 to 4 of ln(3t+1)

Lösung

\int_{0}^{4} ln (3 t + 1) d t

\frac{1}{3} (13 ln (13) - 12)

Dezimale

7.11478 \dots

Schritte zur Lösung

\int_{0}^{4} ln (3 t + 1) d t

Wende U-Substitution an

= \int_{1}^{13} ln (u) \frac{1}{3} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{1}^{13} ln (u) d u

Wende die partielle Integration an

= \frac{1}{3} [u ln (u) - \int 1 d u]_{1}^{13}

\int 1 d u = u

= \frac{1}{3} [u ln (u) - u]_{1}^{13}

Berechne die Grenzen:

13 ln (13) - 12

= \frac{1}{3} (13 ln (13) - 12)

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