integral from 0 to pi/4 of xarctan(x)

Lösung

\int_{0}^{\frac{π}{4}} x arctan (x) d x

\frac{arctan ( \frac{π}{4} ) ( π ^{2} + 16 ) - 4 π}{32}

Dezimale

0.14552 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{4}} x arctan (x) d x

Wende die partielle Integration an

= [\frac{1}{2} (x^{2} arctan (x) - 2 \cdot \int \frac{x ^{2}}{2 ( x ^{2} + 1 )} d x)]_{0}^{\frac{π}{4}}

\int \frac{x ^{2}}{2 ( x ^{2} + 1 )} d x = \frac{1}{2} (- arctan (x) + x)

= [\frac{1}{2} (x^{2} arctan (x) - 2 \cdot \frac{1}{2} (- arctan (x) + x))]_{0}^{\frac{π}{4}}

Vereinfache

[\frac{1}{2} (x^{2} arctan (x) - 2 \cdot \frac{1}{2} (- arctan (x) + x))]_{0}^{\frac{π}{4}} : [\frac{1}{2} (x^{2} arctan (x) + arctan (x) - x)]_{0}^{\frac{π}{4}}

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

Berechne die Grenzen:

\frac{arctan ( \frac{π}{4} ) ( π ^{2} + 16 ) - 4 π}{32}

= \frac{arctan ( \frac{π}{4} ) ( π ^{2} + 16 ) - 4 π}{32}

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