integral from 1 to 3 of (t^6)/(1+t^4)

Lösung

\int_{1}^{3} \frac{t ^{6}}{1 + t ^{4}} d t

\frac{26}{3} - \frac{- ln ( 7 - 2 2 ) + 2 arctan ( 3 2 + 1 ) - 2 arctan ( 2 + 1 ) + ln ( 7 + 2 2 ) + 2 arctan ( 3 2 - 1 ) - 2 arctan ( 2 - 1 )}{4 2}

Dezimale

8.13220 \dots

Schritte zur Lösung

\int_{1}^{3} \frac{t ^{6}}{1 + t ^{4}} d t

Lange Division

\frac{t ^{6}}{1 + t ^{4}} : t^{2} - \frac{t ^{2}}{t ^{4} + 1}

= \int_{1}^{3} t^{2} - \frac{t ^{2}}{t ^{4} + 1} d t

Wende die Summenregel an:

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

= \int_{1}^{3} t^{2} d t - \int_{1}^{3} \frac{t ^{2}}{t ^{4} + 1} d t

\int_{1}^{3} t^{2} d t = \frac{26}{3}

\int_{1}^{3} \frac{t ^{2}}{t ^{4} + 1} d t = \frac{- ln ( 7 - 2 2 ) + 2 arctan ( 3 2 + 1 ) - 2 arctan ( 2 + 1 ) + ln ( 7 + 2 2 ) + 2 arctan ( 3 2 - 1 ) - 2 arctan ( 2 - 1 )}{4 2}

= \frac{26}{3} - \frac{- ln ( 7 - 2 2 ) + 2 arctan ( 3 2 + 1 ) - 2 arctan ( 2 + 1 ) + ln ( 7 + 2 2 ) + 2 arctan ( 3 2 - 1 ) - 2 arctan ( 2 - 1 )}{4 2}

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