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Beliebt Rechnen >

integral of 15xln(1+x)

Lösung

\int 15 x ln (1 + x) d x

Lösung

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

Schritte zur Lösung

\int 15 x ln (1 + x) d x

Entferne die Konstante:

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

= 15 \cdot \int x ln (1 + x) d x

Wende U-Substitution an

= 15 \cdot \int (u - 1) ln (u) d u

Wende die partielle Integration an

= 15 (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

= 15 (ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u))

Setze in

u = 1 + x

ein

= 15 (ln (1 + x) (\frac{( 1 + x ) ^{2}}{2} - (1 + x)) - (\frac{( 1 + x ) ^{2}}{4} - (1 + x)))

Vereinfache

15 (ln (1 + x) (\frac{( 1 + x ) ^{2}}{2} - (1 + x)) - (\frac{( 1 + x ) ^{2}}{4} - (1 + x))) : 15 (\frac{1}{2} ln (1 + x) (1 + x)^{2} - ln (1 + x) - x ln (1 + x) - \frac{1}{4} (1 + x)^{2} + 1 + x)

= 15 (\frac{1}{2} ln (1 + x) (1 + x)^{2} - ln (1 + x) - x ln (1 + x) - \frac{1}{4} (1 + x)^{2} + 1 + x)

Füge eine Konstante zur Lösung hinzu

= 15 (\frac{1}{2} ln (1 + x) (1 + x)^{2} - ln (1 + x) - x ln (1 + x) - \frac{1}{4} (1 + x)^{2} + 1 + x) + C

Graph

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