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

Lösung

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

- \frac{π}{4} + \frac{π ^{2}}{16} - \frac{ln ( 5 ) + 5 arctan ^{2} ( 2 )}{2} + \frac{1}{2} ln (2) + 2 arctan (2)

Dezimale

- 1.47684 \dots

Schritte zur Lösung

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

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

Berechne die Grenzen:

- \frac{π}{4} + \frac{π ^{2}}{16} - \frac{ln ( 5 ) + 5 arctan ^{2} ( 2 )}{2} + \frac{1}{2} ln (2) + 2 arctan (2)

= - \frac{π}{4} + \frac{π ^{2}}{16} - \frac{ln ( 5 ) + 5 arctan ^{2} ( 2 )}{2} + \frac{1}{2} ln (2) + 2 arctan (2)

\int_{1}^{2} \frac{1}{x ( ln ( x ) ) ^{2}} d x

\int_{0}^{\frac{π}{4}} tan (x) d x

\int_{0}^{5} \frac{x}{3 x ^{2} + 6} d x

\int_{0}^{\frac{π}{3}} (sin (x)) d x

\int_{0}^{1} sin (y^{2}) d y