integral of 2x^3ln(x^2+1)

Lösung

\int 2 x^{3} ln (x^{2} + 1) d x

\frac{1}{2} x^{4} ln (x^{2} + 1) - \frac{x ^{4}}{4} + \frac{x ^{2}}{2} - \frac{1}{2} ln x^{2} + 1 + C

Schritte zur Lösung

\int 2 x^{3} ln (x^{2} + 1) d x

Entferne die Konstante:

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

= 2 \cdot \int x^{3} ln (x^{2} + 1) d x

Wende die partielle Integration an

= 2 (\frac{1}{4} x^{4} ln (x^{2} + 1) - \int \frac{x ^{5}}{2 ( x ^{2} + 1 )} d x)

\int \frac{x ^{5}}{2 ( x ^{2} + 1 )} d x = \frac{1}{2} (\frac{x ^{4}}{4} - \frac{x ^{2}}{2} + \frac{1}{2} ln x^{2} + 1)

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

Vereinfache

2 (\frac{1}{4} x^{4} ln (x^{2} + 1) - \frac{1}{2} (\frac{x ^{4}}{4} - \frac{x ^{2}}{2} + \frac{1}{2} ln x^{2} + 1)) : \frac{1}{2} x^{4} ln (x^{2} + 1) - \frac{x ^{4}}{4} + \frac{x ^{2}}{2} - \frac{1}{2} ln x^{2} + 1

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} x^{4} ln (x^{2} + 1) - \frac{x ^{4}}{4} + \frac{x ^{2}}{2} - \frac{1}{2} ln x^{2} + 1 + C

\int (3 x^{2} - 5)^{4} x d x

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