integral of ((4x^3-3x^2-8x))/(6x)

解

\int \frac{( 4 x ^{3} - 3 x ^{2} - 8 x )}{6 x} d x

\frac{1}{6} (\frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2} - 8 x) + C

解答ステップ

\int \frac{4 x ^{3} - 3 x ^{2} - 8 x}{6 x} d x

定数を除く:

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

= \frac{1}{6} \cdot \int \frac{4 x ^{3} - 3 x ^{2} - 8 x}{x} d x

簡素化

\frac{4 x ^{3} - 3 x ^{2} - 8 x}{x} : 4 x^{2} - 3 x - 8

= \frac{1}{6} \cdot \int 4 x^{2} - 3 x - 8 d x

総和規則を適用する:

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

= \frac{1}{6} (\int 4 x^{2} d x - \int 3 x d x - \int 8 d x)

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

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

\int 8 d x = 8 x

= \frac{1}{6} (\frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2} - 8 x)

定数を解答に追加する

= \frac{1}{6} (\frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2} - 8 x) + C