integral from 0 to 2 of 1/(x^3+1)

Lösung

\int_{0}^{2} \frac{1}{x ^{3} + 1} d x

\frac{ln ( 3 ) + 3 π}{6}

Dezimale

1.09000 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{1}{x ^{3} + 1} d x

Ermittle den Partialbruch von

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

= \int_{0}^{2} \frac{1}{3 ( x + 1 )} + \frac{- x + 2}{3 ( x ^{2} - x + 1 )} 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

= \int_{0}^{2} \frac{1}{3 ( x + 1 )} d x + \int_{0}^{2} \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x

\int_{0}^{2} \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln (3)

\int_{0}^{2} \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x = \frac{- ln ( 3 ) + 3 π}{6}

= \frac{1}{3} ln (3) + \frac{- ln ( 3 ) + 3 π}{6}

Vereinfache

\frac{1}{3} ln (3) + \frac{- ln ( 3 ) + 3 π}{6} : \frac{ln ( 3 ) + 3 π}{6}

= \frac{ln ( 3 ) + 3 π}{6}

\int_{2}^{4} \frac{x}{2} d x

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

\int_{0}^{4} x e^{4 x} d x

\int_{0}^{8} 2 x x^{2} + 36 d x