integral of (x+4)(3x-5)

Lösung

\int (x + 4) (3 x - 5) d x

x^{3} + \frac{7 x ^{2}}{2} - 20 x + C

Schritte zur Lösung

\int (x + 4) (3 x - 5) d x

Multipliziere aus

(x + 4) (3 x - 5) : 3 x^{2} + 7 x - 20

= \int 3 x^{2} + 7 x - 20 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 3 x^{2} d x + \int 7 x d x - \int 20 d x

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

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 20 d x = 20 x

= x^{3} + \frac{7 x ^{2}}{2} - 20 x

Füge eine Konstante zur Lösung hinzu

= x^{3} + \frac{7 x ^{2}}{2} - 20 x + C

\int \frac{- y}{x ^{2}} d x

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