integral of x^3ln|1+x|

Lösung

\int x^{3} ln ∣ 1 + x ∣ d x

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

Schritte zur Lösung

\int x^{3} ln ∣ 1 + x ∣ d x

Wende die partielle Integration an

= \frac{1}{4} x^{4} ln ∣ 1 + x ∣ - \int \frac{x ^{4}}{4 ( 1 + x )} d x

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

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

Füge eine Konstante zur Lösung hinzu

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

x \to - 1 lim (\frac{x - 3}{x + 1})

d er i v a t i v e f (x) = \frac{x ^{3} + 2 x ^{2} + 3 x + 4}{x}

\frac{d}{d x} (9 - x)

d er i v a t i v e 4 π x^{2}

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n}