integral of ln(sqrt(x+1))

Lösung

\int ln (x + 1) d x

\frac{1}{2} (x ln (x + 1) + ln (x + 1) - x - 1) + C

Schritte zur Lösung

\int ln (x + 1) d x

Vereinfache

ln (x + 1) : \frac{1}{2} ln (x + 1)

= \int \frac{1}{2} ln (x + 1) d x

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int ln (x + 1) d x

Wende U-Substitution an

= \frac{1}{2} \cdot \int ln (u) d u

Wende die partielle Integration an

= \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} (u ln (u) - u)

Setze in

u = x + 1

ein

= \frac{1}{2} ((x + 1) ln (x + 1) - (x + 1))

Multipliziere aus

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

= \frac{1}{2} (x ln (x + 1) + ln (x + 1) - x - 1)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (x ln (x + 1) + ln (x + 1) - x - 1) + C

\int (3 x^{2} - 1)^{2} d x

\int \frac{1 + x ^{3}}{x ^{2}} d x

\int π tan (π x) d x

\int \frac{( 3 ln ( x ) - 1 ) ^{4}}{x} d x

\int (\frac{1}{e ^{x} + 1}) d x