integral from 1 to e^5 of x^4ln(x)

Lösung

\int_{1}^{e^{5}} x^{4} ln (x) d x

\frac{24 e ^{25} + 1}{25}

Dezimale

69124703363.93034

Schritte zur Lösung

\int_{1}^{e^{5}} x^{4} ln (x) d x

Wende die partielle Integration an

= [\frac{1}{5} (x^{5} ln (x) - 5 \cdot \int \frac{x ^{4}}{5} d x)]_{1}^{e^{5}}

\int \frac{x ^{4}}{5} d x = \frac{x ^{5}}{25}

= [\frac{1}{5} (x^{5} ln (x) - 5 \cdot \frac{x ^{5}}{25})]_{1}^{e^{5}}

5 \cdot \frac{x ^{5}}{25} = \frac{x ^{5}}{5}

= [\frac{1}{5} (x^{5} ln (x) - \frac{x ^{5}}{5})]_{1}^{e^{5}}

Berechne die Grenzen:

\frac{24 e ^{25}}{25} + \frac{1}{25}

= \frac{24 e ^{25}}{25} + \frac{1}{25}

Vereinfache

= \frac{24 e ^{25} + 1}{25}

d er i v a t i v e y = \frac{x ^{2} - 1}{x ^{2} + 1}

\int_{- 1}^{2} (3 x + 1) d x

in v erse l a pl a ce \frac{x ^{2} + 1}{x ^{2} - x - 6}

d er i v a t i v e f (x) = 8 sin^{6} (x)

x \to 4 + lim (\frac{1}{∣ 4 - x ∣})