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

Lösung

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

ln (2) - \frac{1}{2}

Dezimale

0.19314 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{1}^{2} ln (u) d u

Wende die partielle Integration an

= \frac{1}{2} [u ln (u) - \int 1 d u]_{1}^{2}

\int 1 d u = u

= \frac{1}{2} [u ln (u) - u]_{1}^{2}

Berechne die Grenzen:

2 ln (2) - 1

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

Vereinfache

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

\int_{0}^{1} 1 - t d t

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