integral from 0 to 1 of arctan(1-x)

Lösung

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

- \frac{1}{2} ln (2) + \frac{π}{4}

Dezimale

0.43882 \dots

Schritte zur Lösung

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

Wende die partielle Integration an

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

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

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

Vereinfache

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

Berechne die Grenzen:

- (\frac{1}{2} ln (2) - \frac{π}{4})

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

Vereinfache

= - \frac{1}{2} ln (2) + \frac{π}{4}

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