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

Lösung

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

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

Dezimale

0.09657 \dots

Schritte zur Lösung

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

Vereinfache

ln (x^{2} + 1) : \frac{1}{2} ln (x^{2} + 1)

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

Entferne die Konstante:

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

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

Wende U-Substitution an

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

Wende die partielle Integration an

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

\int 1 d u = u

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

Vereinfache

\frac{1}{2} \cdot \frac{1}{2} [u ln (u) - u]_{1}^{2} : \frac{1}{4} [u ln (u) - u]_{1}^{2}

= \frac{1}{4} [u ln (u) - u]_{1}^{2}

Berechne die Grenzen:

2 ln (2) - 1

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

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