integral of (3x+4)^2

Lösung

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

3 x^{3} + 12 x^{2} + 16 x + C

Schritte zur Lösung

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

Multipliziere aus

(3 x + 4)^{2} : 9 x^{2} + 24 x + 16

= \int 9 x^{2} + 24 x + 16 d x

Wende die Summenregel an:

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

= \int 9 x^{2} d x + \int 24 x d x + \int 16 d x

\int 9 x^{2} d x = 3 x^{3}

\int 24 x d x = 12 x^{2}

\int 16 d x = 16 x

= 3 x^{3} + 12 x^{2} + 16 x

Füge eine Konstante zur Lösung hinzu

= 3 x^{3} + 12 x^{2} + 16 x + C

x \to 2 lim (\frac{∣ x - 2 ∣}{2 - x})

x \to 0 lim (- 2 x^{2})

in v erse l a pl a ce \frac{2}{s ( s + 2 ) ( s + 3 )}

n = 1 \sum \infty (- \frac{2}{3})^{n - 1}

\int_{3}^{9} \frac{1}{x ^{2} - 1} d x