integral from 1 to 2 of 7/(x^3+4x)

Lösung

\int_{1}^{2} \frac{7}{x ^{3} + 4 x} d x

7 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

Dezimale

0.80175 \dots

Schritte zur Lösung

\int_{1}^{2} \frac{7}{x ^{3} + 4 x} d x

Entferne die Konstante:

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

= 7 \cdot \int_{1}^{2} \frac{1}{x ^{3} + 4 x} d x

Ermittle den Partialbruch von

\frac{1}{x ^{3} + 4 x} : \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )}

= 7 \cdot \int_{1}^{2} \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 (\int_{1}^{2} \frac{1}{4 x} d x - \int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x)

\int_{1}^{2} \frac{1}{4 x} d x = \frac{1}{4} ln (2)

\int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} (3 ln (2) - ln (5))

= 7 (\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)))

Füge

\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5))

zusammen:

\frac{- ln ( 2 ) + ln ( 5 )}{8}

= 7 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

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