integral from 0 to 2 of x^2ln(x^2+1)

Lösung

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

\frac{8}{3} ln (5) - \frac{2 ( 3 arctan ( 2 ) + 2 )}{9}

Dezimale

3.10929 \dots

Schritte zur Lösung

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

Wende die partielle Integration an

= [\frac{1}{3} x^{3} ln (x^{2} + 1) - \int \frac{2 x ^{4}}{3 ( x ^{2} + 1 )} d x]_{0}^{2}

\int \frac{2 x ^{4}}{3 ( x ^{2} + 1 )} d x = \frac{2}{3} (\frac{x ^{3}}{3} - x + arctan (x))

= [\frac{1}{3} x^{3} ln (x^{2} + 1) - \frac{2}{3} (\frac{x ^{3}}{3} - x + arctan (x))]_{0}^{2}

Berechne die Grenzen:

\frac{8}{3} ln (5) - \frac{2 ( 3 arctan ( 2 ) + 2 )}{9}

= \frac{8}{3} ln (5) - \frac{2 ( 3 arctan ( 2 ) + 2 )}{9}

\int_{0}^{1} \frac{1}{4} (2 x + y) d y

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