integral of (4x+3)^2

Lösung

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

\frac{16 x ^{3}}{3} + 12 x^{2} + 9 x + C

Schritte zur Lösung

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

Multipliziere aus

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

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

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

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

\int 9 d x = 9 x

= \frac{16 x ^{3}}{3} + 12 x^{2} + 9 x

Füge eine Konstante zur Lösung hinzu

= \frac{16 x ^{3}}{3} + 12 x^{2} + 9 x + C

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