integral of 4(x^2+3x+7)(2x+3)

Lösung

\int 4 (x^{2} + 3 x + 7) (2 x + 3) d x

4 (\frac{x ^{4}}{2} + 3 x^{3} + \frac{23 x ^{2}}{2} + 21 x) + C

Schritte zur Lösung

\int 4 (x^{2} + 3 x + 7) (2 x + 3) d x

Entferne die Konstante:

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

= 4 \cdot \int (x^{2} + 3 x + 7) (2 x + 3) d x

Multipliziere aus

(x^{2} + 3 x + 7) (2 x + 3) : 2 x^{3} + 9 x^{2} + 23 x + 21

= 4 \cdot \int 2 x^{3} + 9 x^{2} + 23 x + 21 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

= 4 (\int 2 x^{3} d x + \int 9 x^{2} d x + \int 23 x d x + \int 21 d x)

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

\int 23 x d x = \frac{23 x ^{2}}{2}

\int 21 d x = 21 x

= 4 (\frac{x ^{4}}{2} + 3 x^{3} + \frac{23 x ^{2}}{2} + 21 x)

Füge eine Konstante zur Lösung hinzu

= 4 (\frac{x ^{4}}{2} + 3 x^{3} + \frac{23 x ^{2}}{2} + 21 x) + C

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