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

Lösung

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

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

Dezimale

1.11072 \dots

Schritte zur Lösung

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

Multipliziere aus

\frac{1 + x ^{2}}{1 + x ^{4}} : \frac{1}{1 + x ^{4}} + \frac{x ^{2}}{1 + x ^{4}}

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

\int_{0}^{1} \frac{1}{1 + x ^{4}} d x = \frac{2 ln ( 2 + 2 ) - 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

\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}

= \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}

Vereinfache

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

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

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