integral of (4x^3-3x^2+6x-1)

解

\int (4 x^{3} - 3 x^{2} + 6 x - 1) d x

x^{4} - x^{3} + 3 x^{2} - x + C

解答ステップ

\int 4 x^{3} - 3 x^{2} + 6 x - 1 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

= \int 4 x^{3} d x - \int 3 x^{2} d x + \int 6 x d x - \int 1 d x

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

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

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

\int 1 d x = x

= x^{4} - x^{3} + 3 x^{2} - x

定数を解答に追加する

= x^{4} - x^{3} + 3 x^{2} - x + C