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Beliebt Rechnen >

integral of ((y)ln(1+y))

Lösung

\int ((y) ln (1 + y)) d y

Lösung

\frac{1}{2} ln (1 + y) (1 + y)^{2} - ln (1 + y) - y ln (1 + y) - \frac{1}{4} (1 + y)^{2} + 1 + y + C

Schritte zur Lösung

\int y ln (1 + y) d y

Wende U-Substitution an

= \int (u - 1) ln (u) d u

Wende die partielle Integration an

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

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)

Setze in

u = 1 + y

ein

= ln (1 + y) (\frac{( 1 + y ) ^{2}}{2} - (1 + y)) - (\frac{( 1 + y ) ^{2}}{4} - (1 + y))

Vereinfache

ln (1 + y) (\frac{( 1 + y ) ^{2}}{2} - (1 + y)) - (\frac{( 1 + y ) ^{2}}{4} - (1 + y)) : \frac{1}{2} ln (1 + y) (1 + y)^{2} - ln (1 + y) - y ln (1 + y) - \frac{1}{4} (1 + y)^{2} + 1 + y

= \frac{1}{2} ln (1 + y) (1 + y)^{2} - ln (1 + y) - y ln (1 + y) - \frac{1}{4} (1 + y)^{2} + 1 + y

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln (1 + y) (1 + y)^{2} - ln (1 + y) - y ln (1 + y) - \frac{1}{4} (1 + y)^{2} + 1 + y + C

Graph

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