integral from 0 to 1 of xarctan^2(x)

Lösung

\int_{0}^{1} x arctan^{2} (x) d x

\frac{1}{2} ln (2) + \frac{π ^{2}}{16} - \frac{π}{4}

Dezimale

0.17802 \dots

Schritte zur Lösung

\int_{0}^{1} x arctan^{2} (x) d x

Wende die partielle Integration an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

\frac{1}{2} ln (2) + \frac{π ^{2}}{16} - \frac{π}{4}

= \frac{1}{2} ln (2) + \frac{π ^{2}}{16} - \frac{π}{4}

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