integral of ln(x^2+1)x

Lösung

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

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

Schritte zur Lösung

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

ein

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

Multipliziere aus

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

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

Füge eine Konstante zur Lösung hinzu

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

\int 6 x^{2} 4^{x^{3}} d x

\int (x + 1) (3 x - 1) d x

\int \frac{x ^{2} - 5}{x + 2} d x

\int \frac{1}{x ^{\frac{1}{3}} + x ^{\frac{2}{3}}} d x

\int 2 x^{- \frac{2}{3}} - 2 x^{\frac{5}{3}} d x