integral of 9(x+1)^2ln(x+1)

Lösung

\int 9 (x + 1)^{2} ln (x + 1) d x

9 (ln (x + 1) (\frac{x ^{3}}{3} + x^{2} + x) - \frac{1}{18} (9 x^{2} + 18 x + 2 (x + 1)^{3} - 6 ln ∣ x + 1 ∣ - 9 (x + 1)^{2} + 9)) + C

Schritte zur Lösung

\int 9 (x + 1)^{2} ln (x + 1) d x

Entferne die Konstante:

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

= 9 \cdot \int (x + 1)^{2} ln (x + 1) d x

Wende die partielle Integration an

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

\int \frac{x ^{3} + 3 x ^{2} + 3 x}{3 ( x + 1 )} d x = \frac{1}{18} (9 x^{2} + 18 x + 2 (x + 1)^{3} - 6 ln ∣ x + 1 ∣ - 9 (x + 1)^{2} + 9)

= 9 (ln (x + 1) (\frac{x ^{3}}{3} + x^{2} + x) - \frac{1}{18} (9 x^{2} + 18 x + 2 (x + 1)^{3} - 6 ln ∣ x + 1 ∣ - 9 (x + 1)^{2} + 9))

Füge eine Konstante zur Lösung hinzu

= 9 (ln (x + 1) (\frac{x ^{3}}{3} + x^{2} + x) - \frac{1}{18} (9 x^{2} + 18 x + 2 (x + 1)^{3} - 6 ln ∣ x + 1 ∣ - 9 (x + 1)^{2} + 9)) + C

x \to - 3 lim (\frac{∣ x + 3 ∣}{x + 3})

\int sin^{5} (5 x) d x

\int_{0}^{\infty} 9 e^{- x} d x

\frac{\partial}{\partial x} ((x - 1)^{2} + x y + (y - 1)^{2})

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