integral of (9x^3-5x^2+3x)/x

解

\int \frac{9 x ^{3} - 5 x ^{2} + 3 x}{x} d x

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

解答ステップ

\int \frac{9 x ^{3} - 5 x ^{2} + 3 x}{x} d x

簡素化

\frac{9 x ^{3} - 5 x ^{2} + 3 x}{x} : 9 x^{2} - 5 x + 3

= \int 9 x^{2} - 5 x + 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 9 x^{2} d x - \int 5 x d x + \int 3 d x

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

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 3 d x = 3 x

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

定数を解答に追加する

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