integral from 0 to x of ln(1+x)

Lösung

\int_{0}^{x} ln (1 + x) d x

x ln (x + 1) - x + ln (x + 1)

Schritte zur Lösung

\int_{0}^{x} ln (1 + x) d x

Wende U-Substitution an

= \int_{1}^{1 + x} ln (u) d u

Wende die partielle Integration an

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

\int 1 d u = u

= [u ln (u) - u]_{1}^{1 + x}

Berechne die Grenzen:

x ln (x + 1) - x + ln (x + 1)

= x ln (x + 1) - x + ln (x + 1)

\int_{0}^{0.3} x ln (1 + x^{3}) d x

\int_{- 2}^{2} x^{5} d x

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

\int_{0}^{5} 6

\int_{1}^{2} (4 x^{3} - 6) d x