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

解

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

3 x + \frac{4}{x} + \frac{1}{x ^{2}} + C

解答ステップ

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

拡張

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

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

\int 3 d x = 3 x

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

\int \frac{2}{x ^{3}} d x = - \frac{1}{x ^{2}}

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

簡素化

= 3 x + \frac{4}{x} + \frac{1}{x ^{2}}

定数を解答に追加する

= 3 x + \frac{4}{x} + \frac{1}{x ^{2}} + C