integral from 0 to x of ln(z+2)

Lösung

\int_{0}^{x} ln (z + 2) d z

x ln (x + 2) + 2 ln (x + 2) - x - 2 ln (2)

Schritte zur Lösung

\int_{0}^{x} ln (z + 2) d z

Wende U-Substitution an

= \int_{2}^{x + 2} ln (u) d u

Wende die partielle Integration an

= [u ln (u) - \int 1 d u]_{2}^{x + 2}

\int 1 d u = u

= [u ln (u) - u]_{2}^{x + 2}

Berechne die Grenzen:

x ln (x + 2) + 2 ln (x + 2) - x - 2 ln (2)

= x ln (x + 2) + 2 ln (x + 2) - x - 2 ln (2)

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