integral from 0 to 0.4 of 1/(1+x^4)

Lösung

\int_{0}^{0.4} \frac{1}{1 + x ^{4}} d x

0.39798 \dots

Schritte zur Lösung

\int_{0}^{0.4} \frac{1}{1 + x ^{4}} d x

Ermittle den Partialbruch von

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

= \int_{0}^{0.4} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} + \frac{- x + 2}{2 2 ( x ^{2} - 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}^{0.4} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x + \int_{0}^{0.4} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x

\int_{0}^{0.4} \frac{x + 2}{2 2 ( x ^{2} + 2 x + 1 )} d x = 0.17317 \dots

\int_{0}^{0.4} \frac{- x + 2}{2 2 ( x ^{2} - 2 x + 1 )} d x = 0.22480 \dots

= 0.17317 \dots + 0.22480 \dots

Vereinfache

= 0.39798 \dots

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