integral from 1 to 4 of 1/(x^3)ln(6x)

Lösung

\int_{1}^{4} \frac{1}{x ^{3}} ln (6 x) d x

\frac{15}{64} + \frac{1}{2} ln (6) - \frac{1}{32} ln (24)

Dezimale

1.03094 \dots

Schritte zur Lösung

\int_{1}^{4} \frac{1}{x ^{3}} ln (6 x) d x

Vereinfache

\frac{1}{x ^{3}} ln (6 x) : \frac{ln ( 6 x )}{x ^{3}}

= \int_{1}^{4} \frac{ln ( 6 x )}{x ^{3}} d x

Wende die partielle Integration an

= [- \frac{ln ( 6 x )}{2 x ^{2}} - \int - \frac{1}{2 x ^{3}} d x]_{1}^{4}

\int - \frac{1}{2 x ^{3}} d x = \frac{1}{4 x ^{2}}

= [- \frac{ln ( 6 x )}{2 x ^{2}} - \frac{1}{4 x ^{2}}]_{1}^{4}

Berechne die Grenzen:

\frac{15}{64} + \frac{1}{2} ln (6) - \frac{1}{32} ln (24)

= \frac{15}{64} + \frac{1}{2} ln (6) - \frac{1}{32} ln (24)

\int_{- \infty}^{1} 6 - x d x

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