integral of (x^3)/(9x^2+16)

Lösung

\int \frac{x ^{3}}{9 x ^{2} + 16} d x

\frac{1}{162} (9 x^{2} + 16 - 16 ln 9 x^{2} + 16) + C

Schritte zur Lösung

\int \frac{x ^{3}}{9 x ^{2} + 16} d x

Wende U-Substitution an

= \int \frac{u - 16}{162 u} d u

Entferne die Konstante:

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

= \frac{1}{162} \cdot \int \frac{u - 16}{u} d u

Multipliziere aus

\frac{u - 16}{u} : 1 - \frac{16}{u}

= \frac{1}{162} \cdot \int 1 - \frac{16}{u} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{162} (\int 1 d u - \int \frac{16}{u} d u)

\int 1 d u = u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= \frac{1}{162} (u - 16 ln ∣ u ∣)

Setze in

u = 9 x^{2} + 16

ein

= \frac{1}{162} (9 x^{2} + 16 - 16 ln 9 x^{2} + 16)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{162} (9 x^{2} + 16 - 16 ln 9 x^{2} + 16) + C

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