integral of (1+x+x^2)ln(x)

Lösung

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

x ln (x) + \frac{1}{2} x^{2} ln (x) + \frac{1}{3} x^{3} ln (x) - x - \frac{x ^{2}}{4} - \frac{x ^{3}}{9} + C

Schritte zur Lösung

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

Wende die partielle Integration an

= ln (x) (x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3}) - \int \frac{1}{6} (6 + 3 x + 2 x^{2}) d x

\int \frac{1}{6} (6 + 3 x + 2 x^{2}) d x = x + \frac{x ^{2}}{4} + \frac{x ^{3}}{9}

= ln (x) (x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3}) - (x + \frac{x ^{2}}{4} + \frac{x ^{3}}{9})

Vereinfache

ln (x) (x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3}) - (x + \frac{x ^{2}}{4} + \frac{x ^{3}}{9}) : x ln (x) + \frac{1}{2} x^{2} ln (x) + \frac{1}{3} x^{3} ln (x) - x - \frac{x ^{2}}{4} - \frac{x ^{3}}{9}

= x ln (x) + \frac{1}{2} x^{2} ln (x) + \frac{1}{3} x^{3} ln (x) - x - \frac{x ^{2}}{4} - \frac{x ^{3}}{9}

Füge eine Konstante zur Lösung hinzu

= x ln (x) + \frac{1}{2} x^{2} ln (x) + \frac{1}{3} x^{3} ln (x) - x - \frac{x ^{2}}{4} - \frac{x ^{3}}{9} + C