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

d er i v a t i v e 5 cos^{2} (x) + 5 sin (x)

f (x) = arcsin (x)

in v erse l a pl a ce \frac{2}{4 s ^{2} - 8 s}

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

s l o p e (0, 2), (- 1, - 1)