integral from 0 to 1 of 4arctan(x^2)

Lösung

\int_{0}^{1} 4 arctan (x^{2}) d x

π - 2 (2 arctan (1 + 2) + 2 arctan (2 - 1) + ln (3 - 22))

Dezimale

1.19161 \dots

Schritte zur Lösung

\int_{0}^{1} 4 arctan (x^{2}) d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{1} arctan (x^{2}) d x

Wende die partielle Integration an

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

\int \frac{2 x ^{2}}{x ^{4} + 1} d x = 2 (- \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)))

= 4 [x arctan (x^{2}) - 2 (- \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)))]_{0}^{1}

Berechne die Grenzen:

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

= 4 \cdot \frac{π - 2 ( 2 arctan ( 1 + 2 ) + 2 arctan ( 2 - 1 ) + ln ( 3 - 2 2 ) )}{4}

Vereinfache

= π - 2 (2 arctan (1 + 2) + 2 arctan (2 - 1) + ln (3 - 22))