integral of xln(1+x^2)

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

= \int \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 ln (u) d u

Wende die partielle Integration an

= \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

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

Setze in

u = 1 + x^{2}

ein

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

Multipliziere aus

(1 + x^{2}) ln (1 + x^{2}) - (1 + x^{2}) : ln (1 + x^{2}) + x^{2} ln (1 + x^{2}) - 1 - x^{2}

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (ln (1 + x^{2}) + x^{2} ln (1 + x^{2}) - 1 - x^{2}) + C

\frac{d y}{d x} = (\frac{y}{x})

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y^{^{'}} = \frac{sin ( x )}{y}

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