integral from 0 to 1 of (x^2)/(1+x^4)

Lösung

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{4}} d x

\frac{- 2 ln ( 2 + 2 ) + 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

Dezimale

0.24374 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{4}} d x

Ermittle den Partialbruch von

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

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

\int_{0}^{1} \frac{x}{2 2 ( x ^{2} + 2 x + 1 )} d x = \frac{1}{2} (\frac{1}{2 2} ln (2 + 2) - \frac{1}{2} (arctan (2 + 1) - \frac{π}{4}))

\int_{0}^{1} \frac{x}{2 2 ( x ^{2} - 2 x + 1 )} d x = \frac{1}{2} (\frac{1}{2 2} ln (2 - 2) + \frac{1}{2} (arctan (2 - 1) + \frac{π}{4}))

= - \frac{1}{2} (\frac{1}{2 2} ln (2 + 2) - \frac{1}{2} (arctan (2 + 1) - \frac{π}{4})) + \frac{1}{2} (\frac{1}{2 2} ln (2 - 2) + \frac{1}{2} (arctan (2 - 1) + \frac{π}{4}))

Vereinfache

- \frac{1}{2} (\frac{1}{2 2} ln (2 + 2) - \frac{1}{2} (arctan (2 + 1) - \frac{π}{4})) + \frac{1}{2} (\frac{1}{2 2} ln (2 - 2) + \frac{1}{2} (arctan (2 - 1) + \frac{π}{4})) : \frac{- 2 ln ( 2 + 2 ) + 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

= \frac{- 2 ln ( 2 + 2 ) + 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

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