integral of x^2ln(1+x)

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende die partielle Integration an

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

\int \frac{1}{3} (u^{2} - 3 u + 3) d u = \frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u

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

Setze in

u = 1 + x

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int 4 x cos (4 x^{2}) d x

\frac{d y}{d x} = \frac{y ^{3}}{3 x - 2}, y (1) = - 1

\int_{- \infty}^{5} \frac{1}{x ^{2} + 1} d x

\int 1 + cos (θ) d θ

t an g e n t f (x) = - 7 x^{4} + 11 x^{2} - 4, at x = 1