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

解

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

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

解答ステップ

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

拡張

\frac{x ^{- 2} - 4}{x ^{3}} : \frac{1}{x ^{5}} - \frac{4}{x ^{3}}

= \int \frac{1}{x ^{5}} - \frac{4}{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 \frac{1}{x ^{5}} d x - \int \frac{4}{x ^{3}} d x

\int \frac{1}{x ^{5}} d x = - \frac{1}{4 x ^{4}}

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

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

簡素化

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

定数を解答に追加する

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