integral from 0 to 1 of arctan(x^3)

Lösung

\int_{0}^{1} arctan (x^{3}) d x

\frac{3 π - 2 π + 2 3 ln ( 2 )}{4 3}

Dezimale

0.22507 \dots

Schritte zur Lösung

\int_{0}^{1} arctan (x^{3}) d x

Wende die partielle Integration an

= [x arctan (x^{3}) - \int \frac{3 x ^{3}}{x ^{6} + 1} d x]_{0}^{1}

\int \frac{3 x ^{3}}{x ^{6} + 1} d x = \frac{1}{4} (- 2 ln x^{2} + 1 + ln 4 x^{4} - 4 x^{2} + 4 + 23 arctan (\frac{2 x ^{2} - 1}{3}))

= [x arctan (x^{3}) - \frac{1}{4} (- 2 ln x^{2} + 1 + ln 4 x^{4} - 4 x^{2} + 4 + 23 arctan (\frac{2 x ^{2} - 1}{3}))]_{0}^{1}

Berechne die Grenzen:

\frac{3 π - 2 π + 2 3 ln ( 2 )}{4 3}

= \frac{3 π - 2 π + 2 3 ln ( 2 )}{4 3}

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