integral of 3x^2(x-2)(x+3)

Lösung

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

3 (\frac{x ^{5}}{5} + \frac{x ^{4}}{4} - 2 x^{3}) + C

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Multipliziere aus

x^{2} (x - 2) (x + 3) : x^{4} + x^{3} - 6 x^{2}

= 3 \cdot \int x^{4} + x^{3} - 6 x^{2} 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

= 3 (\int x^{4} d x + \int x^{3} d x - \int 6 x^{2} d x)

\int x^{4} d x = \frac{x ^{5}}{5}

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

\int 6 x^{2} d x = 2 x^{3}

= 3 (\frac{x ^{5}}{5} + \frac{x ^{4}}{4} - 2 x^{3})

Füge eine Konstante zur Lösung hinzu

= 3 (\frac{x ^{5}}{5} + \frac{x ^{4}}{4} - 2 x^{3}) + C

\int_{1}^{2} 3 \frac{ln ( x )}{x ^{2}} d x

\frac{d}{d x} (x d x)

\frac{d}{d x} ((1 + x)^{- 3})

\int (t + \frac{1}{t})^{\frac{3}{2}} (\frac{t ^{2} - 1}{t ^{2}}) d t

\int \frac{1}{e ^{2 y} - 1} d y