integral of (3x-h)^3

解

\int (3 x - h)^{3} d x

\frac{27 x ^{4}}{4} - 9 h x^{3} + \frac{9 h ^{2} x ^{2}}{2} - h^{3} x + C

解答ステップ

\int (3 x - h)^{3} d x

拡張

(3 x - h)^{3} : 27 x^{3} - 27 h x^{2} + 9 h^{2} x - h^{3}

= \int 27 x^{3} - 27 h x^{2} + 9 h^{2} x - h^{3} 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 27 x^{3} d x - \int 27 h x^{2} d x + \int 9 h^{2} x d x - \int h^{3} d x

\int 27 x^{3} d x = \frac{27 x ^{4}}{4}

\int 27 h x^{2} d x = 9 h x^{3}

\int 9 h^{2} x d x = \frac{9 h ^{2} x ^{2}}{2}

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

= \frac{27 x ^{4}}{4} - 9 h x^{3} + \frac{9 h ^{2} x ^{2}}{2} - h^{3} x

定数を解答に追加する

= \frac{27 x ^{4}}{4} - 9 h x^{3} + \frac{9 h ^{2} x ^{2}}{2} - h^{3} x + C